4. The batteries shown in the figure have negligibly small internal resistors. A

Question

4. The batteries shown in the figure have negligibly small internal resistors. Assuming that and , find the current through a. the resistor, b. the resistor, c. and the battery.

4. The batteries shown in the figure have negligibly small internal resistors. Assuming that 20.0 , find the current through 10.0 V and R = +30.0 a. b- c. the 30.0 resistor, the 20.0 resistor, and the 10.0 V battery. 5.00 V The resistance between terminals a and b in the figure is 75 . If the resistors labeled R have the same value, determine R. 5. 120 40 2 5.0

Explanation / Answer

4. current through battery e is i1 (upward) and current through 30 ohm is i2 (downward)

andcurrent through R and 5 V is i3 upward.

KVL in left side loop:
e - 30i2 = 0

i2 = 10/30 = 0.33 A ........Ans(a)

applying KVL in right side loop:

5 - Ri3 - 30i2 = 0

5 - (20i3) - 10 =0

i3 = - 0.25 A ..........ans(b)

c) applying KCL:

i1 + i3 = i2

i1 - 0.25 = 033

i1 = 0.58 A .....Ans

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R and 5 ohm are in series so their equivalent will be R'

R' = R + 5

now R' , 40 ohm and 120 ohm are in parallel connection,

1/R" = 1/R' + 1/40 + 1/120

R" = 30(R +5) / (30 + R + 5) = 30(R +5) / (35 + R)

now R" and R are in series.

Req = R" + R

75 = R + [ (30 (R +5) / (R + 35)]

75(R + 35) = R^2 + 35R + 30R + 150

R^2 - 10R - 2475 = 0

R = 55 ohm ....Ans

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