Question

A 0.200 H inductor is connected in series with a 88.0 Ω resistor and an ac source. The voltage across the inductor is vL=−(12.0V)sin[(487rad/s)t].

Required:
Derive an expression for the voltage vR across the resistor. Express your answer in terms of the variables L, R, VL (amplitude of the voltage across the inductor), ω, and t .

Answer 1

The expression for the voltage
`(\(v_R\))` across the resistor is given by
`\(v_R = -R \omega \cos(\omega t) - L \omega^2 \sin(\omega t)\)`, where R is the resistance, L is the inductance,
`\(\omega\)` is the angular frequency, and t is time.

Step-by-step explanation:

In this circuit, the voltage across the inductor
`(\(v_L\))` is given as
`\(v_L = -VL \sin(\omega t)\)`, where VL is the amplitude of the voltage across the inductor,
`\(\omega\)` is the angular frequency, and t is time.

The voltage across the resistor
`(\(v_R\))` can be determined by applying Kirchhoff's voltage law in the series circuit. The total voltage across the series combination of the resistor and inductor is the sum of the voltage across the resistor
`(\(v_R\))` and the voltage across the inductor
`(\(v_L\))`. Mathematically,
`\(v_R + v_L = 0\)`, as the net voltage in a closed loop is zero. Substituting the given values and expressions, we get
`\(-R \omega \cos(\omega t) - L \omega^2 \sin(\omega t) + VL \sin(\omega t) = 0\)`.

Solving this equation for
`\(v_R\)`, we arrive at the final expression
`\(v_R = -R \omega \cos(\omega t) - L \omega^2 \sin(\omega t)\)`. This expression captures the time-dependent behavior of the voltage across the resistor in the given series circuit. The cosine and sine terms represent the phase relationship between the voltage and current in the circuit, incorporating the effects of resistance R and inductance L.