Question

If the open circuit voltage of a circuit containing ideal sources and resistors is measured at 8 , while the current through the short circuit across the circuit is 200 , what would be the power absorbed by a 60 resistor placed across the terminals

Answer 1

Answer:

The power absorbed by the 60 ohm resistor is 1.064 W

Explanation:

When there's no load connected to the source (open circuit) its voltage output is ideal, since there won't be any voltage drop across it's internal resistance. When there's a shor circuit, the only load is the internal resistance of the source, so we can use Ohm's law to compute the internal resistance, as shown bellow:

Rinternal = Vopenload/Ishortcircuit

Rinternal = 8/200 = 0.04 Ohm

When we connect a load to this source, the total load we'll be the external resistance plus the internal resistance. We can now compute the curent flow when there's a 60 Ohm resistance connected to the terminals of the source by using Ohm's law again:

I = Vsource/(Rexternal + Rinternal)

I = 8/(60 + 0.04) = 8/(60.04) = 0.1332 A

The absorbed power is the product of the voltage across the terminals of the resistor and the current that goes through it. The current is the one we calculated above and the voltage across it's terminals is given by  I*R, so the power output is:

P = I*Vresistor

P = I*(I*R)

P = R*I^2 = 60*(0.1332)^2  = 1.064 W

Answer 2

Given Information:

Voltage = 8 V

Resistance = 60 Ω

Required Information:

Power absorbed by 60 Ω resistor = ?

Answer:

Power absorbed by 60 Ω resistor = 1.06 Watts

Explanation:

It is given in the question that the circuit contains ideal source.

Ideal Voltage Source:

An ideal voltage source is the one which doesn't not have any internal resistance connected in series with it. therefore, there wont be any internal voltage drop in the source.

Power Absorbed by 60 Ω Resistor:

The power absorbed by the 60 Ω resistor placed across the terminals of voltage source is given by

P = I²R

Where the current I flowing in the circuit is given by

I = V/R

I = 8/60

I = 0.133 A

P = (0.133)²60

P = 1.06 Watts