Which statement is true?

1 answer:

7 0

Statement A is true.

When two resistors are connected in series, they result in an equivalent resistance greater than their individual values. A larger resistance causes lesser current to flow.

When two resistors are connected in parallel, they result in an equivalent resistance lesser than their individual values. A lesser resistance causes more current to flow.

So statement A is the correct answer.
When two resistors are connected in series, there is less total current in the circuit than if the two resistors were connected in parallel.

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You charge an initially uncharged 89.9-mf capacitor through a 30.5-ω resistor by means of a 9.00-v battery having negligible int

blsea [12.9K]

1) The differential equation that models the RC circuit is :

(d/dt)V_capacitor + (V_capacitor/RC) = (V_source/RC)

Where the time constant of the circuit is defined by the product of R*C

Time constant = T = R*C = (30.5 ohms) * (89.9-mf) = 2.742 s

2) Charge of the capacitor 1.57 time constants

1.57*(2.742) = 4.3048 s

The solution of the differential equation is

V_capac (t) = (V_capac(0) - V_capac(∞))e ^(-t /T) + V_capac(∞)

Since the capacitor is initially uncharged V_capac(0) = 0

And the maximun Voltage the capacitor will have in this configuration is the voltage of the battery V_capac(∞) = 9V

This means,

V_capac (t) = (-9V)e ^(-t /T) + 9V

The charge in a capacitor is defined as Q = C*V

Where C is the capacitance and V is the Voltage across

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /T) + 9V

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s) + 9V

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s) + 9V = -1.87V +9V

V_capac (4.3048 s) = 7.1275 V

Q (4.3048 s) = 89.9mF*(7.1275V) = 0.6407 C

3) The charge after a very long time refers to the maximum charge the capacitor will hold in this circuit. This occurs when the voltage accross its terminals is equal to the voltage of the battery = 9V

Q (∞) = 89.9mF*(9V) = 0.8091 C

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