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dc pandey electricity and magnetism solution

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<ul><li><p>Solutionsof</p><p>th thLesson 20 to 25</p><p>Electricity &amp; Magnetism</p><p>By DC Pandey</p></li><li><p>Introductory Exercise 20.11. i q</p><p>t= , here q e= , t r</p><p>v=</p><p>2 pi</p><p> i evr</p><p>=</p><p>2pi </p><p> = </p><p>1.6 10 2.2 102 3.14 5 10</p><p>19 6</p><p>11</p><p> = 1.12 10 3 A = 1.12 mA</p><p>2. No. of at oms in 63.45 g of Cu = 6.023 1023 No. of atoms in 1 cm3 (8.89 g) of Cu</p><p> = 6.023 1063.54</p><p>8.8923</p><p> = 8.43 1022</p><p>As one conduction electron is present peratoms,n = 8.43 1022 cm3 or 8.43 m 1028 3</p><p>As i neAvd= v i</p><p>neAd=</p><p> = </p><p>2.08.43 10 1.6 10 3.14</p><p>(0.5 10 )</p><p>28 19</p><p>3 2</p><p> = 1.88 10 6 ms1</p><p>3. Yes. As cur rent al ways flows in the di rec tion ofelec tric field.</p><p>4. False. In the ab sence of po ten tial dif fer ence,elec trons passes ran dom mo tion.</p><p>5. Cur rent due to both pos i tive and neg a tiveions is from left to right, hence, there is a net cur rent from left to right.</p><p>6. i t= +10 4 dqdt</p><p>t= +10 4</p><p> dq t dtq0 0</p><p>10 10 4 = +( ) q t t= + =[ ]10 2 3002 010 C</p><p>Introductory Exercise 20.21. R L</p><p>A=</p><p> = </p><p>1.723.14 2.05</p><p>10 35</p><p>210</p><p>8</p><p>32</p><p> = 0.57 2. (a) J E=</p><p> i JA EA= =</p><p>=</p><p>0.492.75</p><p>3 14 042 1010</p><p>3 2</p><p>8. ( . )</p><p> = 9.87 A(b) V EL= = =0.49 5.8812 V(c) R V</p><p>i= = =</p><p>5.889.87</p><p>0.6 </p><p>3. Let us con sider the con duc tor to be made upof a num ber of el e men tary discs. Thecon duc tor is sup posed to be ex tended to form a com plete cone and the ver tex O of the cone</p><p>Current Electricity20</p></li><li><p>is taken as or i gin with the con duc tor placedalong x-axis with its two ends at x r= and x l r= + . Let be the semi-ver ti cal an gle ofthe cone.Consider an elementary disc of thickness dxat a distance x from origin.Resistance of this disc,</p><p>dR dxA</p><p>= </p><p>If y be the radius of this disc, thenA y= pi 2</p><p>But y x= tan dR dx</p><p>x= </p><p>pi 2 2tan Resistance of conductor</p><p>R dR dxxr</p><p>l r= = + pi 2 2tan</p><p> Rx r</p><p>l r= </p><p>+pi tan2</p><p>1</p><p> = +</p><p>pi tan2</p><p>1 1r l r</p><p> R lr l r</p><p>=</p><p>+</p><p>pi ( ) tan2</p><p>But, r atan = ( ) tanr l b+ = R l</p><p>ab=</p><p>pi</p><p>4. True. </p><p>=</p><p>1</p><p> = =1 1</p><p>5. R RCu Fe=4.1 3.9Cu Fe( ) ( )1 1+ = + T T</p><p>4.1 4.0[ ( )]1 10 203+ T= + 3.9 5.0[ ( )]1 10 203 T</p><p>4.1 16.4+ 10 203( )T= + 3.9 19.5 10 203( )T</p><p> 3.1 0.2 =10 203( )TT =</p><p> 20</p><p>10 30.2</p><p>3.1 </p><p> = 64.5 CT = 84.5 C</p><p>Introductory Exercise 20.31. Potential difference across both the re sis tors </p><p>is 10 V.</p><p>Hence, i1102</p><p>5= = A</p><p>and i2104</p><p>= = 2.5 A</p><p>2. As A is grounded, VA = 0</p><p>V VB A= + =2 2 VV VC A= + =5 5 V</p><p> V VD C= + =10 15 V</p><p> i V VC B1 13= = A</p><p>and i V VD A2 2152</p><p>=</p><p>= = 7.5 A</p><p>3. Cur rent in the given loop isi E= + 15</p><p>8</p><p> V E i E EAB = = +</p><p> =2 2</p><p>158</p><p>0</p><p> E = 5 V 4. Ef fec tive emf,</p><p>E = =8 1 2 1 6 VEffective resistance of circuit</p><p>3 </p><p>i1 i2</p><p>42 10 V</p><p>i2</p><p>i11 B</p><p>2A</p><p>2 V10 V5 V</p><p>C</p><p>D</p><p>i1R1</p><p>R2 R3 10 Vi3i2</p><p>10 V</p></li><li><p> R R r= +external 10 = + =2 10 1 12 </p><p> i ER</p><p>= = =</p><p>612</p><p>0.5 A</p><p>5. As R R2 3= and V V1 2=Potential difference across R1 is zero.Hence, current through R i1 1 0 =and current through R2 i V</p><p>R21</p><p>2=</p><p> = =1010</p><p>1 A</p><p>6. i ER r</p><p>=</p><p>+</p><p>Also, V E ir= i E V</p><p>r=</p><p>Introductory Exercise 20.41.</p><p>Applying KCL at junction Ri = + =1 2 3 A</p><p>V V VST RU QP= =Taking V VST RU=</p><p>6 1= EE = 5 V</p><p>And fromV VST QP=</p><p>6 12= +irr</p><p>i=</p><p>= =</p><p>12 6 63</p><p>2 </p><p>2. Power de liv ered by the 12 V power sup ply,P Vi1 = = =12 3 36 W</p><p>and power dissipated in 3 resister,P i R3 3</p><p>23</p><p>22 3 12= = = W</p><p>Introductory Exercise 20.5</p><p>1. E</p><p>Er</p><p>Er</p><p>Er</p><p>r r r</p><p>=</p><p>+ +</p><p>+ +=</p><p>+ +</p><p>+ +</p><p>1</p><p>1</p><p>2</p><p>2</p><p>3</p><p>3</p><p>1 2 3</p><p>1 1 1</p><p>101</p><p>42</p><p>62</p><p>11</p><p>12</p><p>12</p><p> = + +10 2 32</p><p> = 7.5 V and 1 1 1 1</p><p>1 2 3r r r r= + +</p><p> = + + =11</p><p>12</p><p>12</p><p>2</p><p> r = 12</p><p> = 0.5 2. i E</p><p>R r=</p><p>+</p><p>Rate of dissipation of energy</p><p>P i R= 2 =+</p><p>E RR r</p><p>2</p><p>2( )For maximum or minimum power</p><p>dPdR</p><p>= 0</p><p> E R r R R rR r</p><p>22</p><p>42 0( ) ( )</p><p>( )+ +</p><p>+</p><p>=</p><p> 4</p><p>Qir12 V</p><p>P</p><p>U</p><p>T 2A</p><p>1AR</p><p>S</p><p>E 1</p><p>3</p><p>E r</p><p>R</p><p>I</p><p>Er</p><p>O R</p><p>I</p><p>Er</p><p>O VE</p><p>(a) (b)</p></li><li><p> E R r r RR r</p><p>24 0</p><p>( )( )( )+ </p><p>+=</p><p>E r R</p><p>R r</p><p>2</p><p>3 0( )</p><p>( )</p><p>+=</p><p> R r=d PdR</p><p>E R r r R R rR r</p><p>2</p><p>22</p><p>3 2</p><p>61 3</p><p>=</p><p>+ +</p><p>+</p><p>( ) ( ) ( )( )( )</p><p> = +</p><p>E r RR r</p><p>2</p><p>44 2( )</p><p>( )</p><p>Clearlyd PdR</p><p>2</p><p>2 is negative at R r= .</p><p>Hence, P is maximum at R r=</p><p>and P E rr r</p><p>Ermax ( )</p><p>=</p><p>+=</p><p>2</p><p>2</p><p>2</p><p>4</p><p>3. When the bat ter ies are con nected in se riesE Eeff = =2 4V, r reff = =2 2 </p><p>For maximum powerR r= =eff 2</p><p>and P Ermax</p><p>( )= =</p><p>=</p><p>eff</p><p>effW</p><p>2 2</p><p>44</p><p>4 22</p><p>4. Ig = 5 mA, G = 1 , V = 5 VR V</p><p>IG</p><p>g= =</p><p>55 10</p><p>13</p><p> = 999 A 999 resistance must be connected inseries with the galvanometer.</p><p>5. G = 100 , ig = 50 A, i = 5 mA</p><p> Si G</p><p>i ig</p><p>g=</p><p>=</p><p>50 10 1005 10 50 10</p><p>6</p><p>3 6</p><p> =</p><p>=</p><p>11</p><p>10.01 0.99</p><p> = 10099</p><p>By connecting a shunt resistance of 10099</p><p>.</p><p>6. i VGg</p><p>=</p><p>and R nVi</p><p>G n Gg</p><p>= = ( )1</p><p>7. V EAB =1516</p><p>Potential gradientk V</p><p>LEAB</p><p>= =</p><p>1516 600</p><p>=</p><p>E640</p><p> V/cm </p><p>(a) E kL2</p><p>= L Ek</p><p>= =</p><p>2320 cm</p><p>(b) V kl= = =E E640</p><p>560 78</p><p>Also, V E ir= E ir E = 7</p><p>8 i E</p><p>r=</p><p>8</p><p>AIEEE CornerSubjective Questions (Level 1)</p><p>1. i qt</p><p>net</p><p>= =</p><p>Given, i = 0.7 , t = 1 s, e = 1.6 C10 19</p><p> n ite</p><p>= = </p><p>0.71.6</p><p>110 19</p><p> = 4.375 108 2. q it= = 3.6 3 3600</p><p>= 38880 C3. (a) q it= = =7.5 337.5 C45</p><p>(b) q ne= n qe</p><p>=</p><p>=</p><p>= </p><p>337.51.6</p><p>2.11 2110</p><p>1019</p><p>4. T rv</p><p>=</p><p>2 pi fT</p><p>vr</p><p>= =</p><p>12 pi</p><p> = </p><p>2.23.14 5.3</p><p>102 10</p><p>6</p><p>11</p><p> = 6.6 1019 s1</p><p> I qT</p><p>ef= =</p><p> = 1.6 6.610 1019 19</p><p> = 10.56 A</p><p>5 </p></li><li><p>5. (a) I t= 55 0.65 2</p><p> I dqdt</p><p>=</p><p> dq Idt= q I dt= q Idt t dt= = 08 208 55( )0.65 = </p><p>5520</p><p>82</p><p>0</p><p>8</p><p>[ ]t t0.65</p><p>= =440 20.8 419.2 C(b) If current is constant I q</p><p>t= = =419.2 52.4 A</p><p>86. i vd</p><p> vv</p><p>ii</p><p>d</p><p>d</p><p>2</p><p>1</p><p>2</p><p>1=</p><p> v ii</p><p>vd d2 12</p><p>1</p><p>46 001 20</p><p>10= = ..</p><p>1.20 </p><p> = 6.00 10 4 ms1 7. v i</p><p>neAd=</p><p>=</p><p> 1</p><p>10 10 1 1028 19 48.5 1.6 = 0.735 10 6 ms1</p><p> = 0.735m/s</p><p> t lvd</p><p>= =</p><p> 10</p><p>10</p><p>3</p><p>60.735 = 1.36 109 s = 43 yr</p><p>8. Dis tance cov ered by one elec tron in 1 s= =1 0.05 0.05 cm</p><p>Number of electrons in 1 cm of wire= 2 1021</p><p> Number of electrons crossing a given areaper second= Number of electrons in 0.05 cm of wire = =0.05 2 10 1021 20</p><p> i qt</p><p>net</p><p>= =</p><p> = = 10 10</p><p>110</p><p>20 191.6 1.6 = 16 A</p><p>9. R LA</p><p>= </p><p>Given, = 0.017 - m </p><p> = 1.7 - m10 8 l = 24.0 m</p><p> A d= = </p><p>pi2 2</p><p>102</p><p>32</p><p>3.14 2.05</p><p> = 3.29 10 6 m2</p><p> R = </p><p>1.7 24.03.29</p><p>1010</p><p>86</p><p> = 0.12 10. R L</p><p>A= </p><p> A LR</p><p>=</p><p>If D is density, then</p><p> m DV DA L D LR</p><p>= = =</p><p> 2</p><p>=</p><p> 8.9 1.72 3.50.125</p><p>10 103 8 2( )</p><p> = 1.5 10 2 kg = 15 g11. At 20C, </p><p>R1 600= , R2 300= At 50C,R R t1 1 11 = +( ) = + = 600 1 30 600( )0.001 1.03 = 618 R R t2 2 21 = +( ) = + =300 1 30 336( )0.004 R R R = + = +1 2 618 336 = 954 </p><p> = </p><p>=</p><p>R RR t</p><p>954 900900 30</p><p> R = + =600 300 900 = 0.002 C 1 </p><p>12. As both the wires are con nected in par al lel, V VAl Cu= i R i RAl Al Cu Cu=</p><p> i Ld</p><p>i LdAl Al</p><p>Al</p><p>AlCu Cu</p><p>Cu</p><p>Cu</p><p>pi</p><p>pi2 2=</p><p> d d i Li LCu Al</p><p>Cu Cu Cu</p><p>Al Al Al=</p><p> = </p><p>1 10 2 63</p><p>3 0.0170.028 7.5</p><p>= 0.569 10 3 m = 0.569 mm.</p><p> 6</p></li><li><p>13. (a) E VL</p><p>= =</p><p>=</p><p>0.938 1.2575 10 2</p><p> V/m</p><p>(b) J E=</p><p> =</p><p>1.254.4 107</p><p> = 2.84 - m10 8 14. (a) J E V</p><p>L= =</p><p>Current density is maximum when L isminimum, ie, L d= , potential differenceshould be applied to faces with dimensions 2 3d d . J V</p><p>dmin.=</p><p>.</p><p>(b) i VR</p><p>VAL</p><p>= =</p><p>Current is maximum when L is minimumand A is maximum.Hence, in this case also, V should be appliedto faces with dimensions 2 3d dand i V d d</p><p>dVd</p><p>max =</p><p>=</p><p>( )( )</p><p>2 3 6 </p><p>.</p><p>15. (a) R LA</p><p>= </p><p> = RAL</p><p>[r d= =2</p><p>1.25 mm = 1.25 10 3 m]</p><p> = 0.104 3.14 1.25</p><p>14.0( )10 3 2</p><p> = 3.64 - m10 7 (b) i V</p><p>RELR</p><p>= = = =1.280.104</p><p>172.3 A14</p><p>(c) i neAvd=v i</p><p>neAd=</p><p>=</p><p> 172.3</p><p>8.5 10 1.6 10 3.14 (1.25 10 )28 19 3 2</p><p>= 2.58 10 3 ms1</p><p>16. For zero ther mal co ef fi cient of resistance, R = 0 R T R TC C Fe Fe + = 0</p><p>RR</p><p>1</p><p>2</p><p>3</p><p>31010</p><p>=</p><p>=</p><p>Fe</p><p>C</p><p>5.00.5</p><p>= 10</p><p> R R1 210=Also, R R1 2 20+ = 10 202 2R R+ =</p><p> R22011</p><p>= = 1.82 </p><p>and R R1 220 20= = 1.82 = 18.2 </p><p>17. The cir cuit can be re drawn as</p><p>Reff 4.8=</p><p>+=</p><p>8 1212 8</p><p> I VR</p><p>= = =</p><p>eff 4.824 5 A</p><p>18. Here, A and C are at same po ten tial and Band D are at same po ten tial,</p><p>Hence, the circuit can be redrawn as</p><p> 1 14</p><p>18</p><p>112</p><p>16R</p><p>= + + +</p><p>=</p><p>+ + +6 3 2 424</p><p> = =1524</p><p>58</p><p> R = 85</p><p> = 1.6 i V</p><p>R= =</p><p>241.6</p><p> = 15 A19. Given cir cuit is sim i lar to that in pre vi ous</p><p>ques tion but 4 re sis tor is re moved. So theef fec tive cir cuit is given by</p><p>7 </p><p>81224 V</p><p>4</p><p>8</p><p>12</p><p>6</p><p>B</p><p>24V</p><p>CD</p><p>A</p><p>24V 84 6</p><p>A,C</p><p>B,D</p><p>12</p></li><li><p>1 18</p><p>112</p><p>16R</p><p>= + +</p><p> 1 3 2 424</p><p>924</p><p>38R</p><p>=</p><p>+ += =</p><p> R = =83</p><p> 2.67</p><p> i VR</p><p>= = =</p><p>24 92.67</p><p>A</p><p>20.</p><p>Wheatstone bridge is bal anced, hence 4 re sis tance con nected be tween B and C bere moved and the ef fec tive cir cuit becomes</p><p>i VR</p><p>= =</p><p>1236 13/</p><p>=</p><p>133</p><p>A </p><p>21. (a) i = ++ +</p><p>=</p><p>12 61 2 3</p><p>3 A</p><p> VG = 0 V VA G= + =12 12 V V VA B = 3 VVB = =12 3 9 V V VB C = 6 V VC = =9 6 3 V V VG D = 6 V, VD = 6 V(b) If 6 V battery is reversed</p><p>i = + +</p><p>=</p><p>12 61 2 3</p><p>1 A</p><p>VG = 0, V vA G = 12 V, VA = 12 V V VA B = 1 V VB = 11 V V VB C = 2 V VC = 9 V V VD G = 6 V VD = 6 V</p><p> 8</p><p>A</p><p>B</p><p>C</p><p>D</p><p>1</p><p>i = 3A</p><p>12 V</p><p>6 V</p><p>G 2</p><p>3</p><p>B</p><p>12V</p><p>CD</p><p>A 6</p><p>2</p><p>6124</p><p>3</p><p> A</p><p>12</p><p>4</p><p>2</p><p>6</p><p>3B12</p><p>D</p><p>C4</p><p>8 6</p><p>A,C</p><p>B,D</p><p>24 V 12</p><p>A</p><p>B</p><p>C</p><p>D</p><p>i = 1A</p><p>12 V</p><p>6 V</p><p>GA</p><p>12</p><p>4</p><p>2</p><p>6</p><p>3B12V</p><p>D</p><p>C</p><p>9 12</p><p>A</p><p>12V 6</p><p>12V36 13</p></li><li><p>22. i =+ +</p><p>=</p><p>2005 10 25</p><p>5 A</p><p>(i) V V3 0 5 25 = V3 125= V(ii) V V0 2 5 10 = </p><p>V2 50= V(iii) V V2 1 5 5 = </p><p>V1 75= V(iv) V3 2 5 35 175 = = V(v) V1 2 5 5 25 = = V(vi) V1 3 200 = V</p><p>23. (a) </p><p> R R R R rv aeff = + +||</p><p> = +</p><p>+ +50 20050 200</p><p>2 1</p><p> = 43 i E</p><p>R= = =</p><p>eff</p><p>4.3 0.1 A43</p><p> Reading of ammeter, i = 0.1 Aand reading of voltmeter = i R Rv( || )</p><p>= =0.1 V40 4(b) </p><p> R R R R ra veff = + +( )||</p><p> = +</p><p>+52 20052 200</p><p>1</p><p> = 42.26 i E</p><p>R= =</p><p>eff0.102 </p><p>Reading of voltmeter V E ir= = 4.3 0.102 1 4.2 Reading of Ammeter,</p><p>i VR Ra</p><p>1 42=</p><p>+= =</p><p>4.2 0.08 A</p><p>24. Con sider the di rec tions of cur rent as shownin fig ure.</p><p>Applying KVL in loop 1, 2 and 3, werespectively get,I I I I1 1 2 16 5 42+ + =( ) 12 6 421 2I I = 2 71 2I I = (i)</p><p>4 6 8 102 2 1 2 3I I I I I+ + + =( ) ( ) 9 3 4 52 1 3I I I + = (ii)</p><p>8 16 42 3 3( )I I I+ + =2 6 12 3I I+ = (iii)</p><p>On solving, we get,I1 = 4.7 A, I2 = 2.4 A, I3 = 0.5 A</p><p>Resistor 5 1 4 6 8 16Current 4.7 A 4.7 A 2.4 A 2.3A 2.9 A 0.5A</p><p>25.</p><p>9 </p><p>r</p><p>A</p><p>V</p><p>Ei</p><p>i1 R</p><p>i2</p><p>42V 5 4</p><p>6</p><p>1</p><p>6</p><p>8</p><p>4VD</p><p>B</p><p>10V</p><p>C</p><p>A</p><p>I1 I2</p><p>I3</p><p>r</p><p>A</p><p>V</p><p>Ei</p><p>i1 R</p><p>Si2</p><p>1</p><p>0</p><p>3</p><p>200 V2</p><p>5</p><p>10</p><p>25</p><p>i1</p><p>VR = 400400</p><p>200100i2</p><p>100</p><p>100</p><p>i 10 V</p><p>i1</p><p>200100</p><p>i2</p><p>100</p><p>100</p><p>i 10 V</p><p>i</p><p>200</p><p>DB</p><p>A C</p></li><li><p>As Wheatstone bridge is balanced, 100resistance between B and D can be removed,ie,</p><p> i i1 210</p><p>3001</p><p>30= = = A</p><p>Hence, reading of voltmeter = Potential difference between B and C = =200 20</p><p>32i V</p><p> = 6.67 V26. (a) (i) When S is open.</p><p>VR</p><p>R RE</p><p>V V1</p><p>1</p><p>1 2</p><p>30005000</p><p>200=+</p><p>= </p><p> = 120 V</p><p> VR</p><p>R REV</p><p>V V2</p><p>2</p><p>1 2</p><p>20005000</p><p>200=+</p><p>= </p><p> = 80 V(ii) When S is closed, </p><p>Now, R1 and V1 are in parallel and theireffective resistance </p><p>RR R</p><p>R RV</p><p>V1</p><p>1</p><p>1</p><p>1</p><p>1</p><p>60005</p><p>1200 =+</p><p>= = </p><p>Similarly,R2 and V2 are in parallel with their effectiveresistance,</p><p>RR R</p><p>R RV</p><p>V2</p><p>2</p><p>2</p><p>2</p><p>2</p><p>60005</p><p>1200 =+</p><p>= = </p><p>As R R1 2 = Hence, reading of V1 = reading of V2</p><p> =+</p><p> =1200</p><p>1200 1200200 100 V</p><p>(b) Current distribution is shown in figure</p><p>i ER R</p><p>=</p><p> + 1 2</p><p> i = =2002400</p><p>112</p><p>A</p><p>iR</p><p>R RiV</p><p>V1</p><p>1</p><p>1</p><p>1</p><p>=</p><p>+ = 3000</p><p>50001</p><p>12</p><p>=</p><p>120</p><p>A</p><p>i RR R</p><p>iV</p><p>21</p><p>1 1</p><p>20005000</p><p>112</p><p>=</p><p>+= </p><p>=</p><p>130</p><p>A</p><p> Current flowing through S i i= = 1 2</p><p>120</p><p>130</p><p> = 160</p><p>A</p><p>27. Ef fec tive emf of 2 V and 6 V bat ter iescon nected in par al lelE E r E r</p><p>r r =</p><p>+</p><p>+=</p><p>+1 2 2 1</p><p>1 2</p><p>2 1 6 11 1</p><p> = 2 Vand r r r</p><p>r r =</p><p>+=</p><p>1 2</p><p>1 2</p><p>12</p><p>= 0.5 </p><p> 10E</p><p>'R2'R1</p><p>R1 S R2</p><p>E</p><p>V1 V2i2 i1</p><p>i2i1i</p><p>V1 V2</p><p>R1 S R2</p><p>E</p><p>V1 V2</p><p>R1 S R2</p><p>E</p><p>i2</p><p>A 100 200</p><p>100 200i1</p><p>10 V</p><p>B C</p><p>D</p></li><li><p>Net emf, E = =4 2 2 V28. (a) </p><p>As E E1 2&gt;Current will flow from B to A.(b) E1 is doing positive work(c) As current flows from B to A throughresistor, B is at higher potential.</p><p>29. i R2 2 5= </p></li><li><p> rR r</p><p>VIR+</p><p>=</p><p> Rr</p><p>IR VV</p><p>=</p><p> = 5 2500 100100</p><p>r = =10012400</p><p>2500 20.16</p><p>37. </p><p>Let R be the resistance of voltmeterAs reading of voltmeter is 30 V,1 1</p><p>4001</p><p>300R+ = R = 1200</p><p>If voltmeter is connected across 300resistor,</p><p>Effective resistance of 300 resistor andvoltmeter</p><p>R = +</p><p>=</p><p>300 1200300 1200</p><p>240</p><p> i =+</p><p>60400 240</p><p> = 60640</p><p>A</p><p> = 332</p><p>A</p><p> Reading of voltmeter, V iR= = 3</p><p>32240</p><p>= 22.5 V38. V R</p><p>R RV2</p><p>1 2=</p><p>+ , </p><p>R rRr R2</p><p>2</p><p>2</p><p>1203</p><p> =</p><p>+=</p><p> = 40 V2</p><p>4060 40</p><p>120=+</p><p> = 48 V</p><p>39. Si</p><p>i iG Rg</p><p>g=</p><p>+( )</p><p> Ri i</p><p>iS Gg</p><p>g=</p><p>=</p><p>20 1010</p><p>203</p><p>3 0.005</p><p> = 79.995 </p><p>40. r L LL</p><p>R= = =1 22</p><p>50.52 0.40.4</p><p>1.5 </p><p>41. Let R be the re sis tance of voltmeter</p><p>R RRe</p><p>= + ++</p><p>3 2 100100</p><p>= ++</p><p>5 100100</p><p>RR</p><p> i RR</p><p>=</p><p>++</p><p>=</p><p>3.4 .045 100</p><p>100</p><p>0</p><p> 0.2 3.4++</p><p>=</p><p>4100</p><p>RR</p><p> R = 400 Reading of voltmeter,</p><p>V i RR</p><p>= +</p><p>100100</p><p> = +</p><p>0.04 100 400100 400</p><p> = 3.2 VIf the voltmeter had been ideal,Reading of voltmeter </p><p>= =100105</p><p>3.4 3.24 V</p><p>42. LL</p><p>RR</p><p>1</p><p>2</p><p>1</p><p>2= </p><p> LL</p><p>1</p><p>1408</p><p>12= (L L1 2 40+ = cm)</p><p> L1 16= cm from A.</p><p> 12</p><p>G BR</p><p>S</p><p>A</p><p>60 V</p><p>V</p><p>300 400</p><p>60 V</p><p>V</p><p>300 400</p><p>3.4 V</p><p>A</p><p>i3</p><p>V</p><p>100</p></li><li><p>43. Si</p><p>i iG Rg</p><p>g=</p><p>+( )</p><p> Ri i</p><p>iS Gg</p><p>g=</p><p>=</p><p> 20 0.0224</p><p>0.02440.0250 9.36</p><p> = 12.94 44. (a) i E</p><p>R rV=</p><p>+ </p><p>V iR ER r</p><p>RVV</p><p>V= = +</p><p>(b) rR rV +</p><p>= =1 1100</p><p>%</p><p>R rV = = 99 99 0.45= 44.55 </p><p>(c) VE</p><p>RR r</p><p>V</p><p>V=</p><p>+</p><p>As RV decreases, V decreases, decreasingaccuracy of voltmeter.</p><p>45. (a) When am me ter is con nectedI E</p><p>R R rA A=</p><p>+ +</p><p>When ammeter is removedI E</p><p>R rR R r</p><p>R rIA A= +</p><p>=</p><p>+ +</p><p>+</p><p>(b) IIA</p><p>= 99%</p><p>R rR R rA</p><p>+</p><p>+ +=</p><p>99100</p><p> R R rA = + = +1</p><p>991</p><p>99( ) (3.8 0.45)</p><p>RA = 0.043 </p><p>(c) As II</p><p>R rR R r</p><p>A</p><p>A=</p><p>+</p><p>+ + , as RA increases, IA</p><p>decreases, decreasing the accuracy ofammeter.</p><p>46. IRmaxmax</p><p>= = =</p><p> 36 152.4</p><p>A</p><p>For the given circuitR R R Re = + =</p><p>12</p><p>32</p><p>Maximum power dissipated by the circuitP I Re =max max2</p><p> = =15 32</p><p>542.4 W</p><p>47. To tal power of the cir cuit, P P P P= + +1 2 3= + +40 60 75= 175 W </p><p>As P VR</p><p>=</p><p>2 R V</p><p>P=</p><p>2</p><p> = =( )120175</p><p>282.3 </p><p>48. Ther mal power gen er ated in the battery</p><p>P i r i E V12</p><p>= = ( ) = 0.6 WPower development in the battery by electric forces</p><p>P IE2 = = 2.6 W49. The given c