An electric circuit is given below. V1 = 1V and Resistance R=1000 Ω. The current through the resistance R is very close to 1 mA and the voltage across point A and B, VAB = 1V. Now the circuit is changed to: where value of V2 = 5V. The internal resistances of both the batteries are 0.1Ω . The current through the resistance R is about:
Calculation:
Equivalent potential
Eeq = \(\frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1}+\frac{1}{r_2}} = \frac{\frac{1}{0.1} + \frac{5}{0.1}}{\frac{1}{0.1}+\frac{1}{0.1}} = 3V\)
Equivalent internal resistance
1/req = 1/0.1 + 1/0.1
req = 0.05 Ω
The total resistance in the circuit, Rtotal = R + r
⇒ Rtotal = 1000 Ω + 0.05 Ω
⇒ Rtotal = 1000.05 Ω
⇒ Current, I = V1 / Rtotal
⇒ I = 3 / 1000.05 Ω
⇒ I ≈ 3 mA
∴ The current through the resistance R in the initial circuit is 3 mA.
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