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Consider the circuit shown in Fig. P25.73. The emf source has negligible internal resistance. The resistors have resistances R1=6.00Ωand R2=4.00Ω. The capacitor has capacitance C=9.00μF. When the capacitor is fully charged, the magnitude of the charge on its plates is Q=36.00μC. Calculate the emfε.

Short Answer

Expert verified

The emf εis 6.67 V

Step by step solution

01

Determine the formula to find the emf

The voltage drop in the circuit is due to the external resistors R1andR2 and from Ohm’s law the emf of the battery is

ε=IR1+R2

Now, the capacitor and resistance R1are in parallel, thus they have the same voltage, thus, the voltage is

V=QC=369V=4V

02

Determine the emf

Now, the current in the circuit is

I=VR1=46I=0.667A

So,

∈=0.667A6+4∈=6.67V

Therefore, the emf εis 6.7 V

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Fig. E25.29.

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