Question

The electrical power \(P\) drawn from a generator by a lightbulb of resistance \(R\) is \(P=V^{2} / R,\) where \(V\) is the line voltage. The resistance of bulb \(\mathrm{B}\) is \(42 \%\) greater than the resistance of bulb A. What is the ratio \(P_{\mathrm{B}} / P_{\mathrm{A}}\) of the power drawn by bulb \(\mathrm{B}\) to the power drawn by bulb \(\mathrm{A}\) if the line voltages are the same?

Short answer

Answer: The ratio of power drawn by bulb B to the power drawn by bulb A is 1/1.42.

Step-by-step solution

Set up expressions for the resistance of bulb B and the power drawn by each bulb

Let's denote the resistance of bulb A as \(R_A\). Then the resistance of bulb B will be \(R_B = R_A + 0.42R_A = 1.42R_A\). Since Power, \(P = \frac{V^2}{R}\), we can write the power drawn by bulb A as \(P_A = \frac{V^2}{R_A}\) and the power drawn by bulb B as \(P_B = \frac{V^2}{R_B}\).

Derive the ratio formula for the powers

We are to find the ratio \(\frac{P_B}{P_A}\). Using the expressions for \(P_A\) and \(P_B\), we can write the ratio as: $$\frac{P_B}{P_A} = \frac{\frac{V^2}{R_B}}{\frac{V^2}{R_A}}$$Since the line voltages are the same, we can cancel out the \(V^2\) term:$$\frac{P_B}{P_A} = \frac{R_A}{R_B}$$Now, substitute the expression for \(R_B = 1.42R_A\): $$\frac{P_B}{P_A} = \frac{R_A}{1.42R_A}$$

Calculate the ratio

Simplify the expression to get the ratio:$$\frac{P_B}{P_A} = \frac{1}{1.42}$$Thus, the ratio of power drawn by bulb B to the power drawn by bulb A is \(\boxed{\frac{1}{1.42}}\).