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In a circuit, a 100.-ohm resistor and a 200.-ohm resistor are connected in parallel to a 10.0-volt battery.

Determine the power dissipated by the 100.-ohm resistor.

Calculate the current in the 200.-ohm resistor.

Respuesta :

Answer:

Explanation:

Let the two resistors be  R1=100Ω  and  R2=200Ω  

The voltage of power source, V =40V  

The total resistance of the circuit is  RT=R1+R2=100+200=300Ω  

Therefore the total current flowing through the circuit is

i=VRT=40300=0.133A  

Therefore the voltage across the two resistors is

V1=iR1=0.133×100=13.33V  

V2=V−V1=40–13.33=26.67V  

Power delivered to the resistor  R1  and  R2  are

P1=V1i=13.33×0.133=1.773watt  

P2=V2i=26.67×0.133=3.547watt

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