Question

Find the current supplied by the battery as shown in the figure.

• A. 4
• B. 5
• C. 54
• D. 3

Answer 1

The correct option is B 5

According to question given,

One end of first group of two resistance 4 and 6 ohms joined with positive terminal of battery and another end at the common potential point similarly one end of other group of 6 and 4 ohms resistance is jointed with negative terminal of battery and another end at a common potential point.

In the above diagram we have reduced circuit,

For parallel combination,

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$

For first group

$\frac{1}{R_1^{\prime }}=\frac{1}{4}+\frac{1}{6}=\frac{5}{12}$

$R_1^{\prime }=\frac{12}{5}$

Similarly

For second group

$R_2^{\prime }=\frac{12}{5}$

Both groups are connected in series

$R_{eq}=R_1^{\prime }+R_2^{\prime }=\frac{24}{5}$

$\text{And we know from ohms law Voltage (V)}=Current\left(I\right)\times Resistance\left(R\right)I=\frac{V}{R}=\frac{24}{24/5}=\frac{24\times 5}{24}=5Amp$