Question

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

• A. 5 J
• B. 10 J
• C. 20 J
• D. 30 J
  

Answer 1

The correct option is C 20 J

Hint : Use the mathematical expression of Joule's Law to evaluate heat dissipated.

Step 1: Find the equivalent resistance of the Circuit.

Given: Two resistors $R_1$ & $R_2$ are connected in series.

So, equivalent resistance can be given as:

$R_{net}$ = $R_1+$$R_2$

$R_{net}$ = $2+$$4=$ $6\Omega$

Step 2 : Find the current through $4\Omega$ resistor

Since both resistors are connected in series so current through

each resistor is same as the current through the circuit.

$Current=\frac{P.D.}{Netresistance}$

$I=\frac{6}{6}=1A$

Step 3: Heat dissipated through $4\Omega$ resistor can be given as :

$H=I^{2}Rt$

where, $H=$ Heat dissipated, $R$= Resistance and $t=$ time in seconds

So, $H=1^2\times 4\times 5=20J$

Final Step : Heat dissipated through $4\Omega$ resistor in $5sec$ is $20J$. Option C is correct.