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To get 5V peak from a 10V peak supply across \$1 \mu F\$ capacitor at 100Hz we need current equal to I = 5V/Xc = 5V/1.592kOhm = 3.1407mA and the voltage drop across resistor is $$V_R = \sqrt{(10V^2 - 5V^2)} = 8.66V peak $$
therefore $$R = \frac{8.66V}{3.1407mA} =2.757k\Omega $$ So yes, your calculations are correct.
And the simulation result look like this [![enter image description here][1]][1]enter image description here

As you can see the peak voltage is 5V but RMS value is around 3.5V [1]: https://i.sstatic.net/6azUQ.png

To get 5V peak from a 10V peak supply across \$1 \mu F\$ capacitor at 100Hz we need current equal to I = 5V/Xc = 5V/1.592kOhm = 3.1407mA and the voltage drop across resistor is $$V_R = \sqrt{(10V^2 - 5V^2)} = 8.66V peak $$
therefore $$R = \frac{8.66V}{3.1407mA} =2.757k\Omega $$ So yes, your calculations are correct.
And the simulation result look like this [![enter image description here][1]][1]

As you can see the peak voltage is 5V but RMS value is around 3.5V [1]: https://i.sstatic.net/6azUQ.png

To get 5V peak from a 10V peak supply across \$1 \mu F\$ capacitor at 100Hz we need current equal to I = 5V/Xc = 5V/1.592kOhm = 3.1407mA and the voltage drop across resistor is $$V_R = \sqrt{(10V^2 - 5V^2)} = 8.66V peak $$
therefore $$R = \frac{8.66V}{3.1407mA} =2.757k\Omega $$ So yes, your calculations are correct.
And the simulation result look like this enter image description here

As you can see the peak voltage is 5V but RMS value is around 3.5V

deleted 5 characters in body
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G36
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To get 5V peak from a 10V peak supply across \$1 \mu F\$ capacitor at 100Hz we need current equal to I = 5V/Xc = 5V/1.592kOhm = 3.1407mA and the voltage drop across resistor is $$V_R = \sqrt{(10V^2 - 5V^2)} = 8.66V peak $$
therefore $$R = \frac{8.66V}{3.1407 \Omega} =2.757k\Omega $$$$R = \frac{8.66V}{3.1407mA} =2.757k\Omega $$ So yes, your calculations are correct.
And the simulation result look like this [![enter image description here][1]][1]

As you can see the peak voltage is 5V but RMS value is around 3.5V [1]: https://i.sstatic.net/6azUQ.png

To get 5V peak from a 10V peak supply across \$1 \mu F\$ capacitor at 100Hz we need current equal to I = 5V/Xc = 5V/1.592kOhm = 3.1407mA and the voltage drop across resistor is $$V_R = \sqrt{(10V^2 - 5V^2)} = 8.66V peak $$
therefore $$R = \frac{8.66V}{3.1407 \Omega} =2.757k\Omega $$ So yes, your calculations are correct.
And the simulation result look like this [![enter image description here][1]][1]

As you can see the peak voltage is 5V but RMS value is around 3.5V [1]: https://i.sstatic.net/6azUQ.png

To get 5V peak from a 10V peak supply across \$1 \mu F\$ capacitor at 100Hz we need current equal to I = 5V/Xc = 5V/1.592kOhm = 3.1407mA and the voltage drop across resistor is $$V_R = \sqrt{(10V^2 - 5V^2)} = 8.66V peak $$
therefore $$R = \frac{8.66V}{3.1407mA} =2.757k\Omega $$ So yes, your calculations are correct.
And the simulation result look like this [![enter image description here][1]][1]

As you can see the peak voltage is 5V but RMS value is around 3.5V [1]: https://i.sstatic.net/6azUQ.png

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G36
G36
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To get 5V peak from a 10V peak supply across 1uF\$1 \mu F\$ capacitor at 100Hz we need Icurrent equal to I = 5V/Xc = 5V/1.592kOhm = 3.1407mA and the voltage drop across resistor is $$V_R = \sqrt{(10V^2 - 5V^2)} = 8.66V peak $$
therefore $$R = \frac{8.66V}{3.1407 \Omega} =2.757k\Omega $$ So yes, your calculationcalculations are correct.
And the simulation result look like this [![enter image description here][1]][1]

As you can see the peak voltage is 5V but RMS value is around 3.5V [1]: https://i.sstatic.net/6azUQ.png

To get 5V peak across 1uF capacitor at 100Hz we need I = 5V/Xc = 5V/1.592kOhm = 3.1407mA and the voltage drop across resistor is $$V_R = \sqrt{(10V^2 - 5V^2)} = 8.66V peak $$
therefore $$R = \frac{8.66V}{3.1407 \Omega} =2.757k\Omega $$ So yes your calculation are correct.
And the simulation result look like this [![enter image description here][1]][1]

As you can see the peak voltage is 5V but RMS value is around 3.5V [1]: https://i.sstatic.net/6azUQ.png

To get 5V peak from a 10V peak supply across \$1 \mu F\$ capacitor at 100Hz we need current equal to I = 5V/Xc = 5V/1.592kOhm = 3.1407mA and the voltage drop across resistor is $$V_R = \sqrt{(10V^2 - 5V^2)} = 8.66V peak $$
therefore $$R = \frac{8.66V}{3.1407 \Omega} =2.757k\Omega $$ So yes, your calculations are correct.
And the simulation result look like this [![enter image description here][1]][1]

As you can see the peak voltage is 5V but RMS value is around 3.5V [1]: https://i.sstatic.net/6azUQ.png

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G36
G36
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