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.
Two cells of emf 1V and 2V and internal resistance 2 \( \Omega \) and 1 \( \Omega \), respectively, are connected in series with an external resistance of 6 \( \Omega \). The total current in the circuit is \( I_1 \). Now the same two cells in parallel configuration are connected to the same external resistance. In this case, the total current drawn is \( I_2 \). The value of \( \left( \frac{I_1}{I_2} \right) \) is \( \frac{x}{3} \). The value of x is 1cm.
In the figure shown below, a resistance of 150.4 $ \Omega $ is connected in series to an ammeter A of resistance 240 $ \Omega $. A shunt resistance of 10 $ \Omega $ is connected in parallel with the ammeter. The reading of the ammeter is ______ mA.