Question

A student found that when a resistor of resistance $3\Omega$ was joined with 3 V battery as per figure shown below, the current flowing through it was 1 A. He then joined another resistor of resistance $6\Omega$ in parallel with a resistor of resistance $3\Omega$. The reading in the ammeter will now be:

(a) 9 A
(b) 1.5 A
(c) 1 A
(d) 6 A

Answer 1

Given, the resistance of the resistors are $R_1=3\Omega$ and $R_2=6\Omega$
Voltage, $V=3volts$
Let the combined resistance of the resistors $R_1$ and $R_2$ be $R$. Since the resistors are connected in series, $\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$
$\Rightarrow R=\frac{R_1{R}_2}{R_1+R_2}$ = $\frac{3\times 6}{3+6}$ $=2\Omega$

We know, $V=IR$
$\Rightarrow I=\frac{V}{R}$ = $\frac{3}{2}=1.5A$