A 8 Ω resistor is in series with parallel combination of two resistors 12 Ω and 6 Ω if the current in 6Ω is 5 ampere determine the total power dissipated in the circuit.
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Voltage drop across 6 ohms = 6 ohms * 5 amp = 30 volts
Current in 12 ohm resistor = voltage drop / resistance
= 30 / 12 = 2.5 amp
By applying kirchoff's current law :
current in 8 ohms resistor = 5 amp + 2.5 amp = 7.5 amp
Power dissipated = i*i* R =>
= 7.5 * 7.5 * 8 + 5 * 5 * 6 + 2.5 * 2.5 * 12
= 675 watts
Voltage drop across 6 ohms = 6 ohms * 5 amp = 30 volts
Current in 12 ohm resistor = voltage drop / resistance
= 30 / 12 = 2.5 amp
By applying kirchoff's current law :
current in 8 ohms resistor = 5 amp + 2.5 amp = 7.5 amp
Power dissipated = i*i* R =>
= 7.5 * 7.5 * 8 + 5 * 5 * 6 + 2.5 * 2.5 * 12
= 675 watts
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