A circuit is constructed with four resistors, one capacitor, one battery and a s

Question

A circuit is constructed with four resistors, one capacitor, one battery and a switch as shown. The values for the resistors are: R_1 - R_2 - 76 ohm, R_3 = 65 ohm and R_4 = 61 ohm. The capacitance is C = 50 mu F and the battery voltage is V = 24V. The switch has been open for a long time when at time t = 0, the switch is closed. What is l_1 (0), the magnitude of the current through the resistor R_1 just after the switch is closed? A You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. What is l_1(-), the magnitude of the current that flows through the resistor R_1 a very long time after the switch has been closed? A You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. What is Q(-), the charge on the capacitor after the switch has been closed for a very long time? mu C You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. Consider the circuit above, with R5 = 51 ohm in series with the capacitor. Once again, the switch has been open for a long time when at time t = 0, the switch is closed. What is l_1(0), the magnitude of the current through the resistor R_1 just after the switch is closed? You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. Continuing with the new circuit, what is Q(-), the charge on the capacitor after the switch has been closed for a very long time? mu C You currently have 0 submissions for this question. Only 10 submission are a You can make 10 more submissions for this question.

Explanation / Answer

1. I1(0) = V/(R1+R4) = 24/(76 +61) =0.175 Amp......Ans.

2. I1() = V/(R1+R2+R3+R4) = 24/(76 +76+65+61) =0.0863 Amp......Ans.

3.

A current loop equation to find the voltage across the capacitor,

Vc First note I1=I4 at t->infinity

V-I1 R1-Vc-I4 R4=0 so Vc=V-I1 R1-I4 R4 = 24-4.96-70(.0919) = 12.607 V

Now use Q=VC Q=VcC= 12.607 V(78 F)x 10-6 =983.346 C

4.

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