An ideal battery with E = 12 V is connected in series to two bulbs with resistance R1= 2 ohm andR2= 4 ohm. what is the current in the circuit and the power dissipation in each bulb ?

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Solution

$N e t r e s i s \tan c e o f t h e c i r c u i t i s g i v e n b y : R = R_{1} + R_{2} = 2 + 4 = 6 o h m E = 12 V o l t c u r r e n t i n t h e c i r c u i t I = E / R I = 12 / 6 = 2 A . p o w e r d i s s i p a t i o n i n b u l b 1 = p_{1} = I^{2} R_{1} = 4 \times 2 = 8 w a t t p_{2} = I^{2} R_{2} = 4 \times 4 = 16 w a t t$

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An ideal battery with E = 12 V is connected in series to two bulbs with resistance R1= 2 ohm andR2= 4 ohm. what is the current in the circuit and the power dissipation in each bulb ?

Net resistance of the circuit is given by:R = R1+R2 = 2+4 = 6 ohmE = 12 Voltcurrent in the circuit I = E/RI = 12/6 = 2 A.power dissipation in bulb 1 = p1 = I2R1=4×2 = 8 wattp2 =I2R2 = 4×4 = 16 watt