step by step please The following circuit contains an ideal 12.0 V battery and 4

Question

step by step please

The following circuit contains an ideal 12.0 V battery and 4 resistors. Assume that all connecting leads have zero resistance and there is no stray resistance in any of the electrical connections. The current in resistor R_1 equals 1.00 A 2.00 A 0.600 mA 1.50 mA 2.00 mA The current in resistor R_2 equals 4/3 mA 2/3 mA 1/3 A 2/3 A 1.00 mA The current in resistor R_3 equals 2/3 mA 1/3 A 4/3 mA 1.00 mA 2/3 A The current in resistor R_4 equals 1/3 A 2/3 A 1.00 mA 2/3 mA 4/3 mA The potential difference across resistor R_1 equals 12.0 V 4.00 V 8.00 V 6.00 V 9.00 V The potential difference across resistor R_3 equals 2.57 V 12.0 V 4.00 V 4.00 V 5.33 V The potential difference across resistor R_4 equals 12.0 V 9.00 V 5.00 V 8.00 V 4.00 V

Explanation / Answer

R2 & R3 are in series

R23 = R2 + R3 = 4 + 2 = 6 k ohms

R23 & R4 are in parallel

R234 = (R23*R4)/(R23+R4) = (6*12)/(6+12) = 4 k ohms

R1 , R234 are in series

Req = R1 + R234 = 6 k ohms

13)

I1 = V/Req = 12/6000 = 2 mA

14)

I2 = V23/R23 = (V-I1*R1)/R23 = (12-4)/6 = 4/3 mA

15)

I3 = I2 = 4/3 mA

16)

I4 = V23/R4 = 2/3 A

17)

v1 = I1*R1 = 4 V

18)

V2 = I2*R2 = 16/3 V

19)

V3 = I3*R3 = 8/3

20)

v4 = I4*R4 = 2/3*12 = 8 V

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