In the circuit on the right, a light bulb, two batteries, and five resistors are

Question

In the circuit on the right, a light bulb, two batteries, and five resistors are wired to form a circuit. For the following question, assume the color codes on resistors are not accurate (that is, use stated values) and assume each battery has 5.75 ? of internal resistance.

If the internal resistance of the light bulb is 531 ?, calculate the power dissipation of the bulb.

In the circuit on the right, a light bulb, two batteries, and five resistors are wired to form a circuit. For the following question, assume the color codes on resistors are not accurate (that is, use stated values) and assume each battery has 5.75 of internal resistance. 2 V 339 589 If the internal resistance of the light bulb is 531 calculate the power dissipation of the bulb. Number 5 V 141 111 141

Explanation / Answer

Combine the 3 resistors in parallel: 111||141||141 = 43.11 approx.
Assume clockwise current i1 in left loop and i2 in right loop
Write equations for left loop
+5 -589i1 + 5.75i2 -2 = 0 => i2 = [656.4i1-7]/3.75
Write equations for right loop
5.75i1 -567.75i2 +2 = 0
Now substitute in equation for i2 and solve for i1
5.75i1 - 567.75*[656.4i1-7]/3.75 +2 = 0 => i1 = 10.85ma
i2 = 3.59ma
So the power dissipation of the bulb is 3.59²*0.429 = 5.542mW = 5.542e-3 W

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