Question

Look at the circuit below. What is the voltage between points B and C?OA. 3 VOB. 0.75 V15001.5 VD

Answer 1

Given:

The potential across the battery is

$V=1.5\text{ V}$

The resistance of each bulb in series is

$R=150\text{ }\Omega$

To find:

The voltage between points B and C

Explanation:

The equivalent resistance of the bulbs is,

$\begin{gathered} R_{eq}=R+R \\ =150+150 \\ =300\text{ }\Omega \end{gathered}$

The current in the circuit is,

$\begin{gathered} I=\frac{V}{R_{eq}} \\ =\frac{1.5}{300} \\ =5\times10^{-3}\text{ A} \end{gathered}$

The voltage drop across B and C is the same as the voltage drop across the first bulb, which is,

$\begin{gathered} V_{BC}=I\times R \\ =5\times10^{-3}\times150 \\ =0.75\text{ V} \end{gathered}$

Hence, the required voltage drop is 0.75 V.