Consider a circuit where a cell of emf \( E_0 \) and internal resistance \( r \) is connected across the terminal A and B as shown in the figure. The value of \( R \) for which the power generated in the circuit is maximum, is given by:
Step 1: The total resistance in the circuit is the sum of the internal resistance \( r \) and the external resistance \( R \).
Step 2: The power generated in the circuit is given by:
\[ P = \frac{E_0^2}{R_{\text{total}}} \times R = \frac{E_0^2}{(R + r)^2} \times R, \]
where \( R_{\text{total}} = R + r \) is the total resistance in the circuit.
Step 3: To find the value of \( R \) that maximizes the power, we take the derivative of the power with respect to \( R \) and set it equal to zero:
\[ \frac{dP}{dR} = \frac{E_0^2 (R + r) - E_0^2 R \times 2(R + r)}{(R + r)^4} = 0. \]
Simplifying, we find:
\[ (R + r) - 2R = 0, \]
\[ R = 3r. \]
Thus, the value of \( R \) for which the power is maximum is \( R = 3r \).
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.
Current passing through a wire as function of time is given as $I(t)=0.02 \mathrm{t}+0.01 \mathrm{~A}$. The charge that will flow through the wire from $t=1 \mathrm{~s}$ to $\mathrm{t}=2 \mathrm{~s}$ is:
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.