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Hashiramasenju
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Homework Statement
https://isaacphysics.org/api/images/content/questions/physics/circuits/capacitors/level4/figures/Circuits_potentiometer_capacitor_otp_2.svg
PQ is a slide-wire of uniform resistance, and J is a moveable contact at some point along it, such that the length from P to J is a known distance l. The total length of PQ is also known to be 1.00m. V=24.0V, C=10.0mF.
Find the energy dissipated. Given that initially S1 is closed and S2 is open, and l=326mm. The circuit is left in this arrangement for long enough that the circuit reaches a steady state. Then S2 is opened and S1 is closed simultaneously.
Homework Equations
E=0.5cv^2
The Attempt at a Solution
So using the above eqn i assumed that the pd across the capacitor will be 24v(because the same voltage flows through the variable resistor branch and the capacitor branch. Now as the current stablies there should be no current flowing through the capacitor. Hence the pd across the capacitor is 24v. Now the total energy of the capacitor = total energy dissipated by the resistor ) then i calculated the energy E=0.5*10*(10^-3)*24^2=2.88
Is this right baecause the answer is different
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