Answer:
The amplitude of voltage source will be 662.799 volt
Step-by-step explanation:
We have given the resistance of the circuit R = 529 ohm
Capacitance
![C=9.8\mu F=9.8* 10^(-6)F](https://img.qammunity.org/2020/formulas/physics/college/dsai02c8049ttrmg38t7vozjibduqv24yt.png)
Frequency f = 60 Hz
Capacitive reactance
![X_C=(1)/(\omega C)=(1)/(2* 3.14* 60* 9.8* 10^(-6))=270.80](https://img.qammunity.org/2020/formulas/physics/college/k7k5ik5z7aoon8z64ab8y7hzif9yj8m8z2.png)
Impedance
![Z=√(R^2+X_C^2)=√(529^2+270.80^2)=594.288](https://img.qammunity.org/2020/formulas/physics/college/gg1adkmaxf25r4fskvd7yvjvypdwlroxhi.png)
Power consumed by resistor = 658 W
We know that
![P=i^2R](https://img.qammunity.org/2020/formulas/physics/college/dedlo37hhjzc48qnmuktqv1b46lsal0k0e.png)
![658=i^2* 529](https://img.qammunity.org/2020/formulas/physics/college/hsdv4fnuupfd45yxg1upegogxicki8ctr0.png)
i = 1.11528 A
We know that voltage = current × impedance = 1.11528×594.288=662.799 volt