To solve this, we’ll address each part of the problem step-by-step.

Given:

• Impedance $Z = 600 - j450 \, \Omega$
• Current $I = 5 \angle 35^\circ \, A$
• Frequency $f = 60 \, \text{Hz}$

(i) Determine the voltage across the circuit

The voltage $V$ across the circuit can be calculated using Ohm's law for AC circuits: $V = I \cdot Z$

To do this, we need to represent both $I$ and $Z$ in polar form.

1. Convert $Z$ to polar form: $|Z| = \sqrt{600^2 + (-450)^2} = \sqrt{360000 + 202500} = \sqrt{562500} = 750 \, \Omega$ $\text{Phase of } Z, \theta_z = \tan^{-1}\left(\frac{-450}{600}\right) = -36.87^\circ$ So, $Z = 750 \angle -36.87^\circ \, \Omega$

2. Use $I = 5 \angle 35^\circ \, A$.

3. Calculate $V$: $V = I \cdot Z = 5 \angle 35^\circ \cdot 750 \angle -36.87^\circ$ Multiply magnitudes and add angles: $|V| = 5 \cdot 750 = 3750 \, \text{V}$ $\text{Angle of } V = 35^\circ - 36.87^\circ = -1.87^\circ$ So, $V = 3750 \angle -1.87^\circ \, \text{V}$

Answer for (i):

The voltage across the circuit is: $V = 3750 \angle -1.87^\circ \, \text{V}$

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(ii) Determine the resistance and capacitance in the circuit

The impedance $Z = R - jX_C$ represents a series circuit with a resistor $R$ and a capacitive reactance $X_C$.

1. Identify the resistance $R$ and reactance $X_C$: From $Z = 600 - j450 \, \Omega$, we see: $R = 600 \, \Omega$ $X_C = 450 \, \Omega$

2. Calculate the capacitance $C$: The capacitive reactance $X_C$ is given by: $X_C = \frac{1}{2 \pi f C}$ Rearranging for $C$: $C = \frac{1}{2 \pi f X_C}$ Substituting $f = 60 \, \text{Hz}$ and $X_C = 450 \, \Omega$: $C = \frac{1}{2 \pi \cdot 60 \cdot 450}$ $C = \frac{1}{169646.0033} \approx 5.9 \, \mu\text{F}$

Answer for (ii):

The values are:

• Resistance $R = 600 \, \Omega$
• Capacitance $C \approx 5.9 \, \mu\text{F}$

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Would you like further details on these calculations?

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Related Questions:

1. How would the impedance change if the frequency were doubled?
2. How would an increase in capacitance affect the impedance and current?
3. How can we calculate the power factor of this circuit?
4. What is the phase difference between the voltage and current in this circuit?
5. How would you calculate the real and reactive power in this AC circuit?

Tip:

Always convert impedances and currents to polar form when using Ohm’s law in AC circuits, as this simplifies the multiplication and angle addition.