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 address this circuit problem, we analyze the behavior of a circuit comprising a voltage source, a resistor, and an inductor. This RL circuit has a current rate of change of 8 A/s when the resistance \( R \) is 12 Ω. Our objective is to determine the current at this specific moment.
The induced electromotive force (emf) in an inductor is defined by the equation:
\(E = -L \frac{dI}{dt}\)
where:
\(E\) is the induced EMF.
\(L\) is the inductance, given as 3 H.
\(\frac{dI}{dt}\) is the rate of change of current, provided as 8 A/s.
Substituting the known values into the emf formula yields:
\(E = -3 \times 8 = -24 \, \text{V}\)
In a steady-state condition, the sum of the voltage across the resistor and the inductor equals the supply voltage:
\(V - IR = L \frac{dI}{dt}\)
Given a supply voltage \( V \) of 12 V, and using the provided values for \( R = 12 \, \Omega \) and \( \frac{dI}{dt} = 8 \, \text{A/s} \), we apply the relation:
\(12 - IR = -24\)
Rearranging this equation to solve for \( IR \):
\(IR = 12 + 24 = 36\)
The current \( I \) can then be calculated as:
\(I = \frac{36}{12} = 3 \, \text{A}\)
Consequently, the current in the circuit at the specified instant is 3 A.
