Question

two resistors of 6 ohm and 9 ohm are connected in series to a 120 V source the Power dissipated by the 6 ohm resistor is

Answer 1

Net resistance = 6+9=15 ohm
Current = V/R = 120/15=8 A
P 6 = i^2*r = 8*8*6=64*6=384 w

Answer 2

Given :

Two resistors of 6 ohm and 9 ohm are connected in series .

Voltage applied , V = 120 V .

To Find :

The power dissipated by the 6 ohm resistor .

Solution :

Current flowing in the circuit is :

$I=\dfrac{120}{6+9}\\\\I=\dfrac{120}{15}\\\\I=8\ A$

Potential difference between 6 ohms resistance is :

$V_6=V(\dfrac{6}{6+9})\\\\V_6=120\times \dfrac{6}{15}\\\\V=48\ V$( Voltage divider rule )

Hence , this is the required solution .

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