Find V1 and V2 in the given circuit. Also calculate I1 and I2 and the power
dissipated in the 12Ω and 40Ω resistors.
Answers
Answer:
V1= [12/(12+10)]*15=90/11 V (works as a potential divider; V1 is voltage across 12 ohm resistor)...
V2=[40/(40+6)]*15=300/23 V (similar explanation as before)
I1 is V1/R so (90/11 V)/12 ohm = 15/22 A
I2 is V2/R so (300/23 V)/40 ohm = 15/46 A
power dissipated is I^2 * R
so power dissipated across 12 ohm resistor is (15/22 A)^2 * 12 ohm
power dissipated across 40 ohm resistor is (15/46 A)^2 * 40 ohm
(could u just calculate the last 2 urself...?:))
hope u understand
Given: R1 = 12Ω
R2 = 40Ω
R3 = 6Ω
R4 = 10Ω
Voltage = 15 V
To find: i₁, i₂, V₁, V₂, Power dissipated in 12Ω and 40Ω resistors.
Solution: Resultant resistance of the circuit will be = 12Ω
The total current in the circuit will be V/R = 15/12A = 5/4A
Voltage across R1 and R3 will be same , so i₁×12 = i₃×6
2i₁= i₃
Total current i = i₁+i₃
5/4 = 3i₁
i₁ = 5/12A
Voltage across R2 and R4 will be same, so i₂×40 = i₄×10
4×i₂ = i₄
Total current i = i₂ + i₄
5/4 = i₂ + 4i₂ = 5i₂
i₂ = 1/4 A
Voltage across R1,
V1 = i₁×R1 = 5/12 × 12 = 5 V
V1 = 5V
Voltage across R2,
V2 = I₂× R2 = 1/4×40 = 10 V
V2 = 10V
Power dissipated in 12 Ω = V²/R = 5²/12 Watt
Power dissipated in 12Ω = 25/12 Watt
Power dissipated in 40 Ω = V²/R = 10²/40 = 5/2 Watt
Power dissipated in 40Ω = 5/2 Watt