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