In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.
To determine the current in the RL circuit at the given instant:
1. Circuit Equation:
The governing equation is:
\[
V = L\frac{di}{dt} + IR
\]
Given values:
- \( V = 12 \) V
- \( L = 3 \) H
- \( R = 12 \) Ω
- \( \frac{di}{dt} = -8 \) A/s (current decreasing)
2. Calculation:
\[
12 = 3(-8) + I(12)
\]
\[
12 = -24 + 12I
\]
\[
36 = 12I
\]
\[
I = 3 \text{ A}
\]
The current at this instant is 3 A.
Two batteries of emf's \(3V \& 6V\) and internal resistances 0.2 Ω \(\&\) 0.4 Ω are connected in parallel. This combination is connected to a 4 Ω resistor. Find:
(i) the equivalent emf of the combination
(ii) the equivalent internal resistance of the combination
(iii) the current drawn from the combination
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: