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The batteries shown in the circuit in Figure 19.57 have negligibly small internal resistances. Find the current through (a) the 30.0$\Omega$ resistor, (b) the 20.0$\Omega$ resistor, and (c) the 10.0 $\mathrm{V}$ battery.

a) 0.333 $\mathrm{A}$b) 0.250 $\mathrm{A}$c) 0.583 $\mathrm{A}$

Physics 102 Electricity and Magnetism

Chapter 19

Current, Resistance, and Direct-Current Circuit

Electric Charge and Electric Field

Gauss's Law

Electric Potential

Capacitance and Dielectrics

Current, Resistance, and Electromotive Force

Direct-Current Circuits

Electromagnetic Induction

University of Michigan - Ann Arbor

Hope College

University of Sheffield

McMaster University

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Here in this diagram, the internal resistance of bats are negligible and we are asked to calculate the amount of curran flowing through 20 rather 30 heraster and the turn volt battery suppose current flowing through 10 volt battery is i 18 point c. I 1 splits into 2 parts i 2 from towards the point e and i 3 flowing through the raster having value of 30 register. So, while using junction rule 8, point c, 8, junction c by using the junction rule that algebric sum of current at a point must be equal to 0 poso. Here i 1 is flowing towards point c there. It will be positive and i 2 and 3 are flowing away from point c. That'S why we will take them as negative plus minus i 3. It must be called to 0. So from here we have. I 1 is equal to i 2 plus i 3. This equation, this equation shows that the current flowing towards a point must be equal to the sum of currents flowing away. From that point, let us see name the sequanow, taking the close look abc d, then back to a by using the voltage rule from a to b, the potential of battery increases. That'S why potential change is positive, plus 10.0 volt from b to c these know any potential draught from c to a the potential draught from the resister is minus. I 3 times resistance is 31 point and from d to e there is no voltage graph. So that must be equal to 0 according to kithov's. Second rule. So from here we have minus, i 3 into 30 is equal to minus 10 volt, or we can write. It is 3 into 30. Is equal to minus 10 is equal to plus 10 volt. All we can write is is 30, i 3 is equal to 10 volt per ohm and volt per oh, is equal to ampere so 30 times r 3 is equal to 10 ampere. Now we can easily put the value of i 3 from equation in this equation. So we have 30 into i 3 is equal to i 1 minus. I 2 so 30 times i 1 minus. I 2 is equal to 10 ampere or we can write it. Is i 1 minus. I 2 is equal to 1 upon 3 ampere. Let this name this equation: 1. Now by taking the close loop b c e, f d, a b c e, f d, a then back to b from e to f the potential graph across the ister 20 homes minus i 2 times 20. Here the potential drop tis negative, that is minus 5.0 volt and from a to b, the potential change of bats positive that plus 10.0 volt. There must be equal to 0 according to kiths watash rule. Now we have minus i 2 times, 20 plus 5. Volt is equal to 0 or we can write it as minus. I, 2 into 20 is equal to minus 5 volt from here we have. I 2 equal to 5 volt upon 20 opt so we get i 2 equal to 1 upon 4 ampere, which is equal to 0.25 ampere. This is the value of i 2. Now, by putting i 2 in equation 1, we can easily calculate i 1 so putting by putting i 2 integration. So we have equation 1. Is i 1 minus? I 2? I 1 minus. I 2 is equal to 1 over 3 ampere. So, by putting i 2 here, i 1 minus- i 2 is 1 over 4 ampere is equal to 1 upon 3 ampere or we can write it as i 1 equal to 1 over 3 ampere plus 1 over 4 ampere. So we have i 1 equal to 7 upon 12 ampere, which is equal to 0.583 ampere. Now, by putting value of i 1 and i 2 in equation a we can easily calculate the value of i 3 point. So by putting i 1 and i 2 in equation, we have 3 equal to i 10.583 ampere minus i 2, that is 0.25 ampere. So we have r 3 equal to 0.333 ampere. Now the current flowing to 20, whom rester is actually i 2, which we have calculated having value of 0.25 ampere and the current flowing through 30 home raise ter. That is, i 3 having value of 0.333 ampere and the current flowing to 10 volt battery is actually i 1, that is having value of 0.583 ampere there times.

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