A circuit is constructed with six resistors and two batteries as shown. The batt

Question

A circuit is constructed with six resistors and two batteries as shown. The battery voltages are V_1 = 18 V and V_2 = 12 V. The positive terminals are indicated with a + sign, The values for the resistors are: R_1 = R_5 = 54 ohm, R_2 = R_6 = 107 ohm R_3 = 44 ohm, and R_4 = 115 ohm. The positive directions for the currents I_1, I_2 and I_3 are indicated by the directions of the arrows. 1) What is V_4, the magnitude of the voltage across the resistor R_4? V Submit You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. 2) What is I_3, the current that flows through the resistor R_3? A positive value for the current is defined to be in the direction of the arrow. A Submit You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. 3) What is I_2, the current that flows through the resistor R_2? A positive value for the current is defined to be in the direction of the arrow. 4) What is I_1, the current that flows through the resistor R_1? A positive value for the current is defined to beta in the direction of the arrow. A Submit You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. 5) What is V(a) - V(b), , the potential difference between the points a and b?

Explanation / Answer

(1)

Let current in V2 branch = i

let it divide into i2 (towards left) and i4 (towards right).

Apply KVL for right loop,

V2 - i4* ( R5 + R4) = 0

12 - i4 * ( 54 + 115) = 0

i4 = 0.071 A

magnitude of the voltage across the resistor R4

V4 = i4 *R4 = 0.071 * 115 = 8.165 V

(2)

Apply KVL for left bigger loop,

V2 - i2*R2 - i1*R1 - i2*R6 = 0

12 - (i2* 107) - (i1*54) - (i2*107) =0

12 - (i1* 54) - (i2*214)= 0 ..... eq(1)

Apply KVL for middle loop,

V2 - i2*R2 - i3*R3 -V1 - i2*R6 =0

12 - (i2*107) - (i3*44) - 18 - ( i2*107) = 0

- 6 - (i2*214) - (i3*44) =0

Put i3 = i2 - i1 into above equation,

- 6 - (i2*214) - ((i2 - i1)*44) =0

- 6 - (i2*258) + (i1*44) =0 .......eq(2)

From equations 1 and 2

(4)

i1 = 0.1876 A

(3)

i2 = .0.00873 A

(2)

i3 = i2 - i1 = 0.1788 A

(5)

V(a) - V(b) = i2*R6 = 0.00873 * 107

V(a) - V(b) = 0.935 V

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