Question

a An AC series circuit is composed of a resistance of 2.1 ohm, inductive reactance of 4 ohm, and a capacitive reactance of 1.5 ohm. If a current of 1 ampere is flowing, what is the applied voltage?​

Answer 1

Answer:

32 V is the applied voltage.

Explanation:

An AC series circuit is composed of a resistance of 20 ohm, inductive reactance of 40 ohm, and a capacitive reactance of 15 ohm. If a current of 1.5 ampere is flowing, 32 V is the applied voltage.

Answer 2

Answer:

The applied voltage is equal to $3.26$ $V$.

Explanation:

• In an AC series LCR circuit, resistance, inductor, and capacitor are connected in series.
• Equal current flows through every component in a series circuit.

The formula for the reactance of an ac series LCR circuit is given by the formula:

$X_{net} =\sqrt{R^{2}+(X_{L}-X_{C})^{2} }$                      

where $X_{net} =$ net reactance of circuit.

            $R=$ resistance value.

            $X_{L} =$ inductive reactance.

            $X_{c}=$ capacitive reactance.

Step 1:

It is given that,  Resistance, R = $2.1$ Ω

                         inductive reactance, $X_{L} =$ $4$ Ω

                         capacitive reactance, $X_{c} = 1.5$ Ω

Net reactance is calculated by substituting the above values in the formula

$X_{net} =\sqrt{2.1^{2}+({4}-1.5)^{2} }$ Ω

$X_{net} =\sqrt{10.66}$  Ω

$X_{net}= 3.26$ Ω

Step 2:

Applying Ohm's Law to calculate voltage:

Voltage, $V=I$ × $X_{net}$

Here, $X_{net} = 3.26$ Ω

So, applied voltage, $V=(1A)*(3.26$ $ohm)$

                                 $V= 3.26$ $V$