Question

The current through the battery is

• A. $1A$
• B. $2A$
• C. $3A$
• D. $1.5A$

Answer 1

Answer: (C)

$3A$

$\frac{1}{R_{total}}=\frac{1}{R_1}+\frac{1}{R_2}$

Therefore, $\frac{1}{R_{total}}=\frac{1}{3}+\frac{1}{6}=\frac{1}{0.5}=2\Omega$.

Hence, their equivalent resistance is $2$ ohms.

When the circuit is closed, the resulting current not only flows through the external circuit, but through the source (cell) itself. The cell have an internal resistance, which causes an internal voltage drop, slightly reducing the voltage across the terminals. The larger the current, the larger the internal voltage drop, and the lower the terminal voltage.
The current in the circuit is calculated from the formula $I=\frac{\epsilon }{R+r}$ where,
$\epsilon$- emf of the cell
$R$- external resistance
$r$- internal resistance of the cell.
When the cell of internal resistance of $3$ ohms is connected to the combined resistance of $2$ ohms the current is given as:
That is, $I=\frac{15}{2+3}=\frac{15}{5}=3A$.

Hence, the current through the circuit is 3 amperes.