# A 210 ohm resistor is connected to an AC source with varepsilon_0 = 9.0 V. (a) What is the peak...

## Question:

A {eq}210 \ \Omega {/eq} resistor is connected to an {eq}AC {/eq} source with {eq}\varepsilon_0 = 9.0 \ V {/eq}.

(a) What is the peak current through the resistor if the emf frequency is {eq}100 \ Hz {/eq}?

(b) What is the peak current through the resistor if the emf frequency is {eq}100 \ kHz {/eq}?

## Peak Voltage of an AC Source:

The peak value of the emf of an AC emf source is the maximum value of the potential provided by the AC source.

The instantaneous value of voltage of an AC source at any instant of time {eq}t {/eq} is represented in sinusoidal form as

{eq}V(t) = V_o \sin(\omega t) {/eq}

• {eq}\omega {/eq} = angular frequency of the AC source = {eq}2\pi f {/eq}.
• {eq}f {/eq} = frequency of the AC source.
• {eq}V_o {/eq} = peak value of the AC source.

It is clear form the expression that {eq}V(t) {/eq} is maximum when {eq}\sin(\omega t) = 1 {/eq}. In that case maximum voltage is {eq}V_o {/eq}.

The potential difference across an electrical element of resistance {eq}R {/eq} is given by the Ohm's law as

{eq}V(t) = I(t) R {/eq}

• {eq}I(t) {/eq} = current passing through the element.

The peak value of current {eq}I_o {/eq} related with the peak value of voltage {eq}V_o {/eq} as

{eq}V_o = I_o R {/eq}.

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Assumptions and Given values:

• {eq}R {/eq} = resistance of the resistor = {eq}210\ \Omega {/eq}.
• {eq}\varepsilon_o {/eq} = peak value of...