find the total resistance,voltage drop across each resistors and power for each resistors.
Answers
Answer:
Current Electricity
Explanation:
A ] Total Resistance
in series
R equivalent = R1 + R2 + R3 = 3 kΩ + 10 kΩ + 5 kΩ = 18 kΩ
B ] Current
Current ( I ) = voltage / Resistance
= 9 / 18 kΩ
= 0.5 × 10-³ A
I = 0.5 mA .
C ] Drop across each resistor
1 ) 3 kΩ
V1 = I × r
= 0.5 × 10-³ × 3 × 10³
= 1.5 Volt
2 ) 10 k Ω
V2 = I × r
= 0.5 × 10-³ × 10 × 10³
= 5 volt .
3 ) 5 k Ω
V3 = I × r
= 0.5 × 10-³ × 5 × 10³
= 2.5 volt.
» [ Now to confirm ; add all drop across resistor to get the voltage of connected battery ] «
★ Vnet = V 1 + V2 + V3 = 1.5 + 5 + 2.5 = 9 volt
D ] Power across each resistor
1 ) 3 k Ω
P1 = V1 × I = 1.5 × 0.5 × 10-³
= 0.75 × 10-³ watt or 750 μ watt
2 ) 10 k Ω
P2 = V2 × I = 5 × 0.5 × 10-³
= 2.5 × 10-³ watt
3 ) 5 k Ω
P3 = V3 × I = 2.5 × 0.5 × 10-³
= 1.25 × 10-³ watt
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Hope it helps you .
Answer:
Explanation:
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