Step 1: The power consumed by the bulb is given by the formula:\( P = \frac{V^2}{R} \) where \( P = 60 \, \text{W} \) and \( V = 120 \, \text{V} \).
Step 2: Rearranging to solve for the resistance of the bulb:\( R_{\text{bulb}} = \frac{V^2}{P} = \frac{120^2}{60} = 240 \, \Omega \)
Step 3: The total resistance required for the correct voltage drop is:\( R_{\text{total}} = \frac{220^2}{60} = 240 \, \Omega \)
Step 4: The series resistance needed is:\( R_{\text{series}} = R_{\text{total}} - R_{\text{bulb}} = 240 \, \Omega - 240 \, \Omega = 200 \, \Omega \)
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.
The graph between variation of resistance of a wire as a function of its diameter keeping other parameters like length and temperature constant is
While determining the coefficient of viscosity of the given liquid, a spherical steel ball sinks by a distance \( x = 0.8 \, \text{m} \). The radius of the ball is \( 2.5 \times 10^{-3} \, \text{m} \). The time taken by the ball to sink in three trials are tabulated as shown: