In a series L-R circuit, connected with a sinusoidal ac source, the maximum potential difference across L and R are respectivaly 3 volts and 4 volts.
At an instant the potemtial difference across resistor is 2 volts. The potential difference in volt, across the inductor at the same instant will be:

Options

A. `3 cos 30^(@)`
B. `3 cos 50^(@)`
C. `6 cos 45^(@)`
D. 6

Answer

To solve the problem step by step, we will analyze the given information and apply the relevant concepts of alternating current in an L-R circuit.

### Step 1: Identify the maximum potential differences
We are given:
- Maximum potential difference across the resistor (V₀_R) = 4 volts
- Maximum potential difference across the inductor (V₀_L) = 3 volts

### Step 2: Write the expressions for instantaneous voltages
In a series L-R circuit, the instantaneous voltage across the resistor (V_R) and the inductor (V_L) can be expressed as:
- \( V_R = V₀_R \cdot \sin(\omega t) \)
- \( V_L = V₀_L \cdot \cos(\omega t) \)

Here, \( \sin(\omega t) \) represents the phase of the voltage across the resistor, and \( \cos(\omega t) \) represents the phase of the voltage across the inductor, which leads the current by 90 degrees.

### Step 3: Use the given instantaneous voltage across the resistor
At a certain instant, the potential difference across the resistor is given as:
- \( V_R = 2 \) volts

Using the expression for \( V_R \):
\[ 2 = 4 \cdot \sin(\omega t) \]
From this, we can solve for \( \sin(\omega t) \):
\[ \sin(\omega t) = \frac{2}{4} = \frac{1}{2} \]

### Step 4: Determine the value of \( \cos(\omega t) \)
Using the Pythagorean identity:
\[ \sin^2(\omega t) + \cos^2(\omega t) = 1 \]
Substituting the value of \( \sin(\omega t) \):
\[ \left(\frac{1}{2}\right)^2 + \cos^2(\omega t) = 1 \]
\[ \frac{1}{4} + \cos^2(\omega t) = 1 \]
\[ \cos^2(\omega t) = 1 - \frac{1}{4} = \frac{3}{4} \]
Taking the square root:
\[ \cos(\omega t) = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

### Step 5: Calculate the instantaneous voltage across the inductor
Now we can find the instantaneous voltage across the inductor using the expression for \( V_L \):
\[ V_L = V₀_L \cdot \cos(\omega t) \]
Substituting the known values:
\[ V_L = 3 \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \text{ volts} \]

### Final Answer
The potential difference across the inductor at that instant is \( \frac{3\sqrt{3}}{2} \) volts.

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