Answer :

The voltage across the capacitor does not change when connected to a resistor after being connected to a battery. So, the given statement is

B. False

The voltage across a capacitor in a DC circuit remains constant once it has been connected to a battery for a long time. So in this case, the voltage across the 10 μF capacitor remains at 6 V even after connecting it to the 5 mΩ resistor.

To determine whether the capacitor was connected to a 6 V battery before being connected to the 5 mΩ resistor, we need to consider the time constant of the RC circuit formed by the capacitor and resistor.

The time constant (τ) of an RC circuit is given by the formula:

τ=RC

Where:

R is the resistance in ohms

C is the capacitance in farads

Given:

Capacitance (C) = 10 μF = 10×10−610×10 −⁶ F

Resistance (R) = 5 mΩ = 5×10−35×10 −³ Ω

Let's calculate the time constant (τ):

τ=(5×10 −³ )×(10×10 −⁶ )

τ=50×10 −⁹

τ=50×10 −⁹

Now, let's find the time it takes for the capacitor to charge to approximately 63.2% of its final voltage. This is the time constant (τ).

If the capacitor was connected to a 6 V battery before, the final voltage across the capacitor (V[tex]_{f}[/tex] ) would be 6 V.

So, approximately after τ seconds, the voltage across the capacitor would reach:

V=V[tex]_{f}[/tex] ×(1−e −[tex]^{} \frac{τ }{t }[/tex])

V=6×(1−[tex]e[/tex]⁻¹ )

V≈6×(1−0.3679)

V≈6×(0.6321)

V≈3.7926

Since the voltage across the capacitor after τ seconds is approximately 3.7926 V, which is less than 6 V, it indicates that the capacitor was not connected to a 6 V battery before being connected to the 5 mΩ resistor.

So, the correct answer is:

B. False

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Rewritten by : Barada