Resistors with resistances of 8 ohms and 12 ohms, connected in parallel, are connected to a current source with

| September 27, 2021 | education

Resistors with resistances of 8 ohms and 12 ohms, connected in parallel, are connected to a current source with an EMF of 12 V and an internal resistance of 1.2 ohms. How much power is dissipated in the high resistance resistor?

R1 = 8 ohms.

R2 = 12 ohms.

r = 1.2 ohms.

EMF = 12 V.

N2 -?

The current power N2 is determined by the formula: N2 = U22 / R2, where U2 is the voltage at the ends of the second resistor, R2 is the resistance of the resistor.

We express the current strength in the entire circuit I by Ohm’s law for a closed loop: I = EMF / (R + r).

R = R1 * R2 / (R1 + R2).

R = 8 ohms * 12 ohms / (8 ohms + 12 ohms) = 4.8 ohms.

I = 12V / (4.8 Ohm + 1.2 Ohm) = 2 A.

U = U1 = U2.

U = I * R.

U = 2 A * 4.8 Ohm = 9.6 V.

N2 = (9.6V) 2 / 12Ω = 7.68W.

Answer: more resistance has a power of N2 = (9.6 V) 2/12 Ohm = 7.68 W.



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