To find the total current drawn from the battery, when two resistors are connected in parallel, we first need to calculate the equivalent resistance of the parallel combination of resistors.
For resistors in parallel, the equivalent resistance \( R_{\text{eq}} \) is given by the formula:
\( \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} \)
Substituting the given resistor values, \( R_1 = 4 \Omega \) and \( R_2 = 6 \Omega \):
\( \frac{1}{R_{\text{eq}}} = \frac{1}{4} + \frac{1}{6} \)
To add these fractions, we find a common denominator, which is 12:
\( \frac{1}{4} = \frac{3}{12}, \quad \frac{1}{6} = \frac{2}{12} \)
\( \frac{1}{R_{\text{eq}}} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12} \)
Thus:
\( R_{\text{eq}} = \frac{12}{5} \Omega = 2.4 \Omega \)
Now, using Ohm's law, the total current \( I \) drawn from the battery is given by:
\( I = \frac{V}{R_{\text{eq}}} \)
where \( V = 12 \text{ V} \):
\( I = \frac{12 \text{ V}}{2.4 \Omega} = 5 \text{ A} \)
Therefore, the total current drawn from the battery is 5 A.
The current passing through the battery in the given circuit, is:
A constant voltage of 50 V is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is :