Question

In an electrical circuit, two resistors of 2 Ω and 4 Ω respectively are connected in series to a 6 V battery. The heat dissipated by the 4 Ω resistor in 5 s will be

• A. 5J
• B. 10J
• C. 20J
• D. 30J

Answer 1

The correct option is C

20J

Step1: Given data

Two resistors $R_1=2\Omega ,R_2=4\Omega$ are connected in series

$Potentialdifference=6V$

1. When the same amount of current passes through the resistors, the circuit is said to be linked in series.
2. The voltage across each resistor in such circuits varies.
3. If any resistor in a series connection breaks or a fault occurs, the entire circuit is shut off.

Step2: Calculating equivalent resistance of the circuit

$$\begin{gathered} R_{net}=R_1+R_2 \\ {R}_{net}=2\Omega +4\Omega \\ {R}_{net}=6\Omega \end{gathered}$$

Step3: Calculating current through $R_2=4\Omega$ resistor

$I=\frac{V}{R}[whereI=current,V=potentialdiffernce,R=resis\tan ce]$

$$\begin{gathered} I=\frac{6}{6}[GivenV=6V] \\ I=1A \end{gathered}$$

Step4: Calculating heat dissipated through $R_2=4\Omega$ resistor

$$\begin{gathered} H=I^{2}Rt[whereH=heatdissipated,I=current,R=resis\tan ce,t=timeinsecond] \\ H={\left(1\right)}^2\times 4\times 5 \\ H=20J \end{gathered}$$

Therefore, heat dissipated through $4\Omega$ resistor in $5sec$ is $20J$.

Hence, option C is the right answer.