A circuit is constructed with six resistors and two batteries as shown. The batt
ID: 1884250 • Letter: A
Question
A circuit is constructed with six resistors and two batteries as shown. The battery voltages are V1 = 18 V and V2 = 12 V. The positive terminals are indicated with a + sign, The values for the resistors are: R1 = R5 = 60 , R2 = R6 = 133 R3 = 49 , and R4 = 131 . The positive directions for the currents I1, I2 and I3 are indicated by the directions of the arrows.
of the arrows.
1)
What is V4, the magnitude of the voltage across the resistor R4?
2)
What is I3, the current that flows through the resistor R3? A positive value for the current is defined to be in the direction of the arrow.
3)
What is I2, the current that flows through the resistor R2? A positive value for the current is defined to be in the direction of the arrow.
4)
What is I1, the current that flows through the resistor R1? A positive value for the current is defined to be in the direction of the arrow.
5)
What is V(a) – V(b), the potential difference between the points a and b?
I R a RbR4Explanation / Answer
1)
By kirchoff's law, we have V2 - i4* ( R5 + R4) =0
12 - i4 * (131+60) = 0
i4 = 0.0628 A
magnitude of the voltage across the resistor R4 ,
V4 = i4 * R4 = 0.0628 * 131
V4 = 8.23 V
2)
By kirchoff's law, we have
V2 - i2*R2 - i1*R1 - i2*R6 = 0
12 - (i2* 133) - (i1*60) - (i2*133) =0
12 - (i1* 60 ) - (i2*266) = 0 ..........(1)
By kirchoff's law, we have
V2 - i2*R2 - i3*R3 -V1 - i2*R6 =0
12 - (i2*133) - (i3*49) - 18 - ( i2*133) = 0
-6 - (i2*266) - (i3*49) =0
we have i3 = i2 - i1 , substituting in the above expression
-6 - (i2*266) - (i2 - i1)*49 = 0
-6 - 315*i2 +49*i1 = 0 ............(2)
Solving equations (1) , (2)
we get i1 = 0.180
i2 = 0.00901
i3 = i2 - i1 = -0.170
i3 = -0.170 A
3)
i2 = 0.00901 A
4)
i1 = 0.180 A
5)
V(a) - V(b) = i2*R6 = 0.00901*133 = 1.198 V
V(a) - V(b) = 1.198 V
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