Question

A resistor $R_1$ dissipates the power $P$ when connected to a certain generator. If the resistor $R_2$ is put in series with $R_1$, the power dissipated by $R_1$

• A. decreases
• B. remains the same
• C. any of the above depending upon
  the relative values of $R_1$ and $R_2$
• D. incareases

Answer 1

The correct option is A decreases

Step 1: Relation o f power with resistance

Power dissipated,

$P=\frac{V^2}{R}$

Initially power dissipated by $R_1$,

$\Rightarrow P=\frac{V^2}{R_1}$

When resistance $R_2$ and $R_1$ are connected in series,

Potential drop across $R_1$,

$V^{{}^{\prime }}V-V_2$

Now, power dissipated by resistor $R_1$ is $P^{{}^{\prime }}$

$\Rightarrow P^{{}^{\prime }}=\frac{V^{{}^{\prime }2}}{R_1}$

Step 2: Comparison of power dissipated

Since $R_1,V,V^{{}^{\prime }}$ and $V_2$ are positive, so

$V^{{}^{\prime }}=V$

$\Rightarrow V^{{}^{\prime }}<V$

$\Rightarrow V^{{}^{\prime }2}<V^2$

$\Rightarrow \frac{V^{{}^{\prime }2}}{R_1}<\frac{V^2}{R_1}$

$\therefore P^{{}^{\prime }}<P$

Thus, power dissipation in second case will be decreased.

Hence, option (a) is correct.