Question:
Consider the circuit diagram shown below, where a battery of 12 V is connected to two resistors R1 and R2 in parallel. The resistance of R1 is 2 Ω and the resistance of R2 is 3 Ω. Determine the following:
Answer:
To determine the equivalent resistance (Req) of the circuit, we can use the formula for resistors in parallel:
1/Req = 1/R1 + 1/R2
Substituting the given values:
1/Req = 1/2 + 1/3
1/Req = 3/6 + 2/6
1/Req = 5/6
Taking the reciprocal of both sides:
Req = 6/5 Ω
Therefore, the equivalent resistance of the circuit is 6/5 Ω.
The current flowing through the battery can be determined using Ohm's Law:
I = V / Req
Substituting the given values:
I = 12 / (6/5)
I = 12 * 5/6
I = 10 A
Therefore, the current flowing through the battery is 10 A.
The current flowing through R1 and R2 can be determined using the formula for resistors in parallel:
I1 = I * (R2 / (R1 + R2))
I1 = 10 * (3 / (2 + 3))
I1 = 10 * (3/5)
I1 = 6 A
Similarly,
I2 = I * (R1 / (R1 + R2))
I2 = 10 * (2 / (2 + 3))
I2 = 10 * (2/5)
I2 = 4 A
Therefore, the current flowing through R1 is 6 A, and the current flowing through R2 is 4 A.
The voltage drop across R1 and R2 can be determined using Ohm's Law:
V1 = R1 * I1
V1 = 2 * 6
V1 = 12 V
Similarly,
V2 = R2 * I2
V2 = 3 * 4
V2 = 12 V
Therefore, the voltage drop across R1 and R2 is 12 V.
The power dissipated by R1 and R2 can be determined using the formula:
P = I^2 * R
P1 = I1^2 * R1
P1 = 6^2 * 2
P1 = 72 W
Similarly,
P2 = I2^2 * R2
P2 = 4^2 * 3
P2 = 48 W
Therefore, the power dissipated by R1 is 72 W, and the power dissipated by R2 is 48 W.
Note: This is a sample question and answer for educational purposes. The values and complexity can be adjusted accordingly for a real exam setting.