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Chapter 4: Q43E (page 876)

An emf source with E = 120 V, a resistor with R = 80.0 鈩�, and a capacitor with C = 4.00 碌F are connected in series. As the capacitor charges, when the current in the resistor is 0.900 A, what is the magnitude of the charge on each plate of the capacitor?

Short Answer

Expert verified

When an emf source with E=120 V and a resistor with 80惟and a capacitor of 4.00渭贵are connected in series and the capacitor gets charges when the current in the resistor is 0.900 A then the magnitude of the charge in each plate capacitor is192渭颁

Step by step solution

01

EMF of the battery

Electromotive force (Emf) is the total amount of electrical potential energy a battery can supply before it is connected to an external circuit

02

Potential difference across the circuit

We know that the charges Q on the plates of the capacitor depend on the voltageVC across both the plates and it is given by:

Q=CVC

Where C is the capacitance. To find the value of Q we need to determine VC. To determine theVCwe will use the loop rule.

蔚-IR-VC=0VC=蔚-IR
03

Calculating the Voltage

Now in the above equation we put the values and get:

VC=蔚-IR=(120V)-(0.900A)(80惟)=48V

Now, putting the values of VCand C into the equation we get:

Q=CVC=(4渭贵)(48V)=192渭颁

Therefore, the magnitude of the charge is192渭颁

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