Current Electricity Question 8
Question 8 - 25 January - Shift 2
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 $6 V$ 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.
$12 V$ Cell 1
Show Answer
Answer: (1)
Solution:
Formula: Combination of Resistances
$E _{\text{eq }}=\frac{\frac{12}{3}-\frac{6}{6}}{\frac{1}{3}+\frac{1}{6}}$
$E _{eq}=6 V$
$r _{\text{eq }}=2 \Omega$
$R=4 \Omega$
So, $i=\frac{6}{2+4}=1 A$
