Question

What is umax,c, the value of the maximum energy stored in the capacitor during one cycle?

Answer 1

1) 0.266 H

2) 0.040 J

3)
`0.308\Omega`

Step-by-step explanation:

The diagram of the circuit is missing: find it in attachment.

1) What is L, the value of the inductance of the circuit?

For an RLC circuit, the resonant angular frequency is given by:

`\omega=(1)/(√(LC))`

where

L is the inductance of the circuit

C is the capacitance of the circuit

In this problem, we have:

`\omega=164 rad/s` is the angular frequency of the generator, at which the circuit is in resonance

`C=140 \mu F = 140\cdot 10^(-6)F` is the capacitance

Therefore, solving for L, we find the inductance of the circuit:

`L=(1)/(\omega^2 C)=(1)/((164)^2(140\cdot 10^(-6)))=0.266 H`

2) What is umax,c, the value of the maximum energy stored in the capacitor during one cycle

The maximum energy stored in a capacitor is given by the equation

`U=(1)/(2)CV^2`

where

C is the capacitance of the capacitor

V is the maximum potential difference across the capacitor

Here we have:

`C=140 \mu F = 140\cdot 10^(-6)F` is the capacitance

`V=24 V` is the maximum voltage across the capacitor, which is the emf of the generator

Substituting, we find:

`U=(1)/(2)(140\cdot 10^(-6))(24)^2=0.040 J`

3) What is R, the value of the resistance of the circuit?

Here we want to find the resistance of the circuit.

The resistance of the circuit can be found by using Ohm's Law:

`V=RI`

where

V is the maximum voltage

R is the resistance

I is the maximum current

Here we have:

V = 24 V is the maximum voltage provided by the generator

I = 0.78 A is the maximum current in the circuit

Solving for R, we find the resistance:

`R=(V)/(I)=(0.24)/(0.78)=0.308\Omega`