Plzz someone solve this
By the way, your question's answer
a) In a series circuit, the same current passes through the resistors, so you can take it to be a constant. Power is hence given by:
P = (I^2)R
I in the circuit = 6/(2+1) = 2 A
So power dissipated in 2 ohm resistor = (2^2).2 = 8 Watts.
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b) Potential difference is constant in a parallel circuit. Hence power dissipated across the 2 ohm resistor is directly given as:
(V^2)/R = (4^2)/2 = 8 Watts.
?
An interesting observation might be that power dissipated in a parallel circuit does not depend on the resistor connected in parallel, while in a series circuit, as current flow is determined by all the resistors, power dissipation is dependent on the other resistors in the circuit.
a) In a series circuit, the same current passes through the resistors, so you can take it to be a constant. Power is hence given by:
P = (I^2)R
I in the circuit = 6/(2+1) = 2 A
So power dissipated in 2 ohm resistor = (2^2).2 = 8 Watts.
?
b) Potential difference is constant in a parallel circuit. Hence power dissipated across the 2 ohm resistor is directly given as:
(V^2)/R = (4^2)/2 = 8 Watts.
?
An interesting observation might be that power dissipated in a parallel circuit does not depend on the resistor connected in parallel, while in a series circuit, as current flow is determined by all the resistors, power dissipation is dependent on the other resistors in the circuit.