Question

The resistor $R_1=3\Omega$ and $R_2=1\Omega$ are connected in parallel to a 20 V battery. Find the heat developed in the resistor $R_1$ in one minute

• A. 600 J
• B. 800 J
• C. 6000 J
• D. 8000 J
• E. 7000 J

Answer 1

Answer: (D)

A circuit with two resistances $3\Omega$ and $1\Omega$ are connected with $20V$ battery is as shown in figure below

Since resistors are connected in parallel, so each branch of the circuit has potential drop of $20V$.
$\therefore$ Current in $R_1$ resistor, $I=\frac{V}{R_1}=\frac{20}{3}A$
$\therefore$ Heat developed in $R_1$ is given by $H=I^2{R}_{1}t$
where, $t=1\min =60s$
$\Rightarrow H=\frac{20}{3}\times \frac{20}{3}\times 3\times 60=8000J$