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Chapter 14: Problem 2063

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

Expert verified
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

01

Identify the given values

We are given the following values: - rms current (I_rms) = 10 A (ampere) - Resistance (R) = 12 Ω (ohm)
02

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: \(I_{peak} = 14.14 \mathrm{~A}\)
03

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: \(V_{max} = 169.68 \mathrm{~V}\)
04

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

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Most popular questions from this chapter

A coil having n turns \(\&\) resistance \(R \Omega\) is connected with a galvanometer of resistance \(4 \mathrm{R} \Omega\). This combination is moved from a magnetic field \(\mathrm{W}_{1} \mathrm{~Wb}\) to $\mathrm{W}_{2} \mathrm{~Wb}\( in \)\mathrm{t}$ second. The induced current in the circuit is.... (a) $-\left[\left\\{\mathrm{W}_{2}-\mathrm{W}_{1}\right\\} /\\{5 \mathrm{Rnt}\\}\right]$ (b) $-\mathrm{n}\left[\left\\{\mathrm{W}_{2}-\mathrm{W}_{1}\right\\} /\\{5 \mathrm{Rt}\\}\right]$ (c) $-\left[\left\\{\mathrm{W}_{2}-\mathrm{W}_{1}\right\\} /\\{\mathrm{Rnt}\\}\right]$ (d) $-\mathrm{n}\left[\left\\{\mathrm{W}_{2}-\mathrm{W}_{1}\right\\} /\\{\mathrm{Rt}\\}\right]$

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