Question 425: Current Electricity

Question

Two cells are connected between points A and B as shown. Cell 1 has emf of 12 V and internal resistance of 3$$\Omega$$. Cell 2 has emf of 6V and internal resistance of 6$$\Omega$$. An external resistor R of 4$$\Omega$$ is connected across A and B. The current flowing through R will be ____________ A.

Accepted Answer

$\mathrm{KCL}$ at $A$ gives
$\frac{6-V_{A}}{4}+\frac{0-V_{A}}{6}+\frac{18-V_{A}}{3}=0$
$V_{A}=10$
So current through $4 \Omega=\frac{10-6}{4}=1 \mathrm{~A}$

Current Electricity

Two cells are connected between points A and B as shown. Cell 1 has emf of 12 V and internal resistance of 3$\Omega. Cell 2 has emf of 6V and internal resistance of 6\Omega. An external resistor R of 4\Omega$ is connected across A and B. The current flowing through R will be ____________ A.

Choose the correct answer: