/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q73P Consider the circuit shown in Fi... [FREE SOLUTION] | 91影视

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Chapter 4: Q73P (page 846)

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

Short Answer

Expert verified

The emf 蔚is 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鈥檚 law the emf of the battery is

蔚=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,

鈭�=0.667A6+4鈭�=6.67V

Therefore, the emf 蔚is 6.7 V

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