Chapter 22, Problem 77A

To determine

Current in resistor.

Answer to Problem 77A

  $I=0.15\text{A}$

Explanation of Solution

Given:

Power rating of resistor, $P=5\text{W}$

Resistance of resistor, $R=220\Omega$

Formula Used:

Power of a resistor in given as,

  $P=I^{2}R$

Where, $P$ is power of resistor, $I$ is current through the resistor and $R$ is resistance of resistor.

Calculation:

On substituting the values in above relation,

  $\begin{array}{c}5=I^2\times 220\\ I^2=\frac{5}{220}\\ I=\sqrt{\frac{5}{220}}\\ =0.15\text{A}\end{array}$

Conclusion:

Therefore, the maximum current allowed through the resistor is $I=0.15\text{A}$