Question

If the current through a resistor is increased by 50%, the increase in power dissipated will be (assume the temperature remains constant)
1. 225 %
2. 200 %
3. 250 %
4. 125 %

Answer 1

The formula that relates power, resistance and the current should be used to solve this problem. As the current is increased by some per cent, so, we will add the original value of the current with this increased value and represent the value in terms of power.
Formula used:
$P={{I}^{2}}R$

Complete answer:
From the given information, we have the data as follows.
The current through the resistor is increased by 50%.
The formula that relates power, resistance and the current is given as follows.
$P={{I}^{2}}R$
Where P is the power, I is the current and R is the resistance.
Now we will compute the value of the current when its value is increased by 50%. Let the original value of the current be ‘I’ and let the final value of the current be ‘I’. So, we have,

& I'=I+50 \% \times I \\\ & \Rightarrow I'=I+\dfrac{50}{100}\times I \\\ & \Rightarrow I'=I+\dfrac{1}{2}I \\\ & \therefore I'=\dfrac{3}{2}I \\\ \end{aligned}$$ Now consider the power dissipated once the current value is increased by 50%. Let the original value of the power be ‘P’ and let the final value of the power be ‘P’’. $$\begin{aligned} & P'=I{{'}^{2}}R \\\ & \Rightarrow P'={{\left( \dfrac{3}{2}I \right)}^{2}}R \\\ & \Rightarrow P'=\dfrac{9}{4}({{I}^{2}}R) \\\ & \therefore P'=\dfrac{9}{4}P \\\ \end{aligned}$$ The increase in the value of the power dissipated is given as follows. $$\begin{aligned} & {{P}_{0}}=\dfrac{P'-P}{P}\times 100 \\\ & \Rightarrow {{P}_{0}}=\dfrac{{}^{9}/{}_{4}P-P}{P}\times 100 \\\ & \Rightarrow {{P}_{0}}=\dfrac{{}^{5}/{}_{4}P}{P}\times 100 \\\ & \therefore {{P}_{0}}= 125 \% \\\ \end{aligned}$$ $$\therefore $$The increase in power dissipated will be 125% (assuming the temperature remains constant), **Thus, option (d) is correct.** **Note:** In this question, we were given with the percentage increase in the value of current. Even, the percentage increase in the value of resistance can be given. As the temperature is assumed to be constant, so, the value of resistance remains the same.