Question 1 The figure shows a circuit consisting of a battery and four resistors

Question

Question 1 The figure shows a circuit consisting of a battery and four resistors. What is the current through and the potential difference across each resistor? R1 = 7.00 N, R2 = 14.00, R3 = 4.00 N, R4 = 4.00 N and V= 34.0 V. Current through R1: Potential difference across R1: Current through R2: Potential difference across R2: Current through Rz: Potential difference across R3: Current through R4: Potential difference across R4: Question 2 If you want to discharge a 8.01 pF capacitor to 5.00% of its initial charge in 2.32 ms, what value resistor will you need? Answer:

Explanation / Answer

1.

Here, R1and R2 are in series and their equivalent resistance Ra= 7+14= 21 ohms, R3 and R4 are in series, so their equivalent resistance Rb= 4+4= 8 ohms.

Now, Ra and Rb are in parallel, so their equivalent is R= RaRb/Ra+Rb= 21*8/21+8= 5.79 ohms.

So, current in the circuit is I= V/R= 34/5.79= 5.87 A

Now, this current is divided into 2 parts, Ia into Ra and Ib into Rb.

So, by current division rule, Ia= I*Rb/ Ra+Rb= 5.87*8/ 21+8 = 1.62 A.So, Ib= I-Ia= 5.87- 1.62= 4.25 A

So, current through R1= Ia= 1.62 A, Potential difference across R1= IaR1= 1.62*7= 11.34 V

Current through R2=Ia= 1.62 A , Potential difference across R2= IaR2= 22.68V

Current through R3= Ib= 4.25A, Potential difference across R3= IbR3= 17V

Current through R4= Ib= 4.25A , Potential difference across R4= IbR4= 17 V

2.

From discharging equation of RC circuit , I= Imax*e^-t/RC

Or, 0.05Imax= Imax*e^-(0.00232/R*8.01*10^-6)

-2.99= -0.00232/R*8.01*10^-6

So, value of resistor you'll need is R= 96.7 ohms

please upvote if understood to appreciate

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