The temperature coefficient of resistance \( \alpha \) quantifies how much the resistance of a material changes with temperature.
The formula for calculating \( \alpha \) is:\( \alpha = \frac{R_2 - R_1}{R_1 (T_2 - T_1)} \)
where \( R_1 \) and \( R_2 \) are the resistances at temperatures
\( T_1 = 100^\circ \text{C} \) and \( T_2 = 400^\circ \text{C} \), respectively.
Using the graph, you can estimate the resistance at these temperatures, and after performing the calculation, you find:
\( \alpha = 3 \times 10^{-7} \, \text{°C}^{-1} \)
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: