Question

An alternating current of \(\mathrm{rms}\) value \(10 \mathrm{~A}\) is passed through a \(12 \Omega\) resistance. The maximum potential difference across the resistor is, (a) \(20 \mathrm{~V}\) (b) \(90 \mathrm{~V}\) (c) \(169.68 \mathrm{~V}\) (d) None of these

Short answer

The maximum potential difference across the resistor is \(169.68 \mathrm{~V}\). To find this, we used the relationship between rms current and peak current and then applied Ohm's Law.

Step-by-step solution

Identify the given values

We are given the following values: - rms current (I_rms) = 10 A (ampere) - Resistance (R) = 12 Ω (ohm)

Find the peak current (I_peak)

To find the peak current, we need to use the relationship between rms current and peak current. The formula to find the peak current is: \(I_{peak} = \sqrt{2} \times I_{rms}\) Substitute the given rms current value: \(I_{peak} = \sqrt{2} \times 10\) Now, calculate the peak current: \(I_{peak} = 14.14 \mathrm{~A}\)

Calculate the maximum potential difference (V_max)

Now we have the peak current (I_peak), and we can use Ohm's Law to calculate the maximum potential difference (V_max) across the resistor. Ohm's Law states that: \(V = I \times R\) Here, V represents the potential difference, I represents the current, and R represents the resistance. Since we want to find the maximum potential difference, we need to use the peak current value instead of the rms current value: \(V_{max} = I_{peak} \times R\) Substitute the values of I_peak and R: \(V_{max} = 14.14 \times 12\) Calculate the maximum potential difference: \(V_{max} = 169.68 \mathrm{~V}\)

Identify the correct answer

Comparing the calculated V_max value with the given options, we can see that the maximum potential difference across the resistor is: (c) 169.68 V