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IV. Numerical Problems · Q9

Q.In the circuit shown, a battery of emf 15 V (internal resistance negligible) is connected through a 100 Ω100\ \Omega resistor carrying current I1I_1, and a battery of emf 9 V is connected through another 100 Ω100\ \Omega resistor carrying current I2I_2; both branches meet at a junction and share a third 100 Ω100\ \Omega resistor R3R_3 carrying current I3=I1+I2I_3 = I_1 + I_2. Calculate the currents I1I_1, I2I_2 and I3I_3 in the circuit.

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Step 1. Take the 15 V battery in series with one 100 Ω100\ \Omega resistor (current I1I_1) and the 9 V battery in series with a second 100 Ω100\ \Omega resistor (current I2I_2), both meeting at a junction and continuing together as I3=I1+I2I_3=I_1+I_2 through the third 100 Ω100\ \Omega resistor, R3R_3, back to both batteries.

Step 2. Applying Kirchhoff's voltage rule to the 15 V loop (through its own 100 Ω100\ \Omega and then through R3R_3): 15=100I1+100I3=100I1+100(I1+I2)=200I1+100I215=100I_1+100I_3=100I_1+100(I_1+I_2)=200I_1+100I_2.

Step 3. Applying Kirchhoff's voltage rule to the 9 V loop (through its own 100 Ω100\ \Omega and then through R3R_3): 9=100I2+100I3=100I1+200I29=100I_2+100I_3=100I_1+200I_2.

Step 4. Solving the two simultaneous equations 15=200I1+100I215=200I_1+100I_2 and 9=100I1+200I29=100I_1+200I_2: multiplying the first by 2 gives 30=400I1+200I230=400I_1+200I_2; subtracting the second gives 21=300I121=300I_1, so I1=0.070I_1=0.070 A (matching the given answer).

Step 5. Substituting back, 15=200(0.070)+100I2=14+100I215=200(0.070)+100I_2=14+100I_2, so I2=0.010I_2=0.010 A in magnitude. …

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