Question

Three bulbs of resistance 100. Ω, 200, Ω and 300 Ω are connected in parallel to a 120. V DC power supply. Draw the diagram and find thea) current in each bulb b) current drawn from the power supplyc) total power drawn power supply d) the net resistance of all bulbs

Answer 1

Let's use the formula for electric current.

`I=(V)/(R)`

Where V is the power supply 120 V, and R is the resistance. Let's find the current in each bulb.

`\begin{gathered} I=(120V)/(100\Omega)=1.20A \\ I=(120V)/(200\Omega)=0.6A \\ I=(120V)/(300\Omega)=0.4A \end{gathered}`

(a) The current in each bulb is 1.20A, 0.6A, and 0.4A, respectively.

(b) (c) The diagram of the circuit is

To find the net resistance, we use the following formula.

`(1)/(R)=(1)/(R_1)+(1)/(R_2)+(1)/(R_3)`

Let's use the given magnitudes.

`\begin{gathered} (1)/(R)=(1)/(100\Omega)+(1)/(200\Omega)+(1)/(300\Omega) \\ (1)/(R)=(6+3+2)/(600\Omega) \\ (1)/(R)=(11)/(600\Omega) \\ R=(600)/(11)\Omega \\ R\approx54.55\Omega \end{gathered}`

Therefore, the net resistance of all bulbs is 54.55 ohms.