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.
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In RL circuits, the current can be determined using the equation \( \epsilon = L \frac{dI}{dt} + IR \). Make sure to substitute the correct values and solve for the unknown current.
To solve this problem, we need to understand the behavior of the circuit which includes a source of voltage, a resistor, and an inductor. The circuit given is an RL circuit, and the rate of change of current is given as 8 A/s when the resistance \( R \) is 12 Ω. We want to find the current at this instant.
The formula for the induced electromotive force (emf) in an inductor is given by the equation:
\(E = -L \frac{dI}{dt}\)
where:
\(E\) is the induced EMF.
\(L\) is the inductance (3 H in this case).
\(\frac{dI}{dt}\) is the rate of change of current (8 A/s).
Plugging the values into the formula, we get:
\(E = -3 \times 8 = -24 \, \text{V}\)
In a steady state, the voltage across the resistor and the inductor equals the supply voltage:
\(V - IR = L \frac{dI}{dt}\)
Given that the supply voltage \( V \) is 12 V, and substituting for \( R = 12 \, \Omega \) and \( \frac{dI}{dt} = 8 \, \text{A/s} \), we use the relation:
\(12 - IR = -24\)
Rearranging the equation gives:
\(IR = 12 + 24 = 36\)
Thus, the current \( I \) can be found as:
\(I = \frac{36}{12} = 3 \, \text{A}\)
Therefore, the current in the circuit at the instant described is 3 A.
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Approach Solution -2
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)
