A direct current of 10 A is superimposed on an alternating current / = 40 cos `omegat` (A) flowing through a wire. The effective value of the resulting current will be

Options

A. `10sqrt(2)` A
B. `20 sqrt(2)` A
C. `20 sqrt(3)` A
D. 30 A

Answer

To find the effective value of the resulting current when a direct current (DC) of 10 A is superimposed on an alternating current (AC) described by \( i = 40 \cos(\omega t) \), we can follow these steps:

### Step 1: Define the total current
The total current \( I(t) \) can be expressed as the sum of the DC current and the AC current:
\[ I(t) = I_{DC} + I_{AC} = 10 + 40 \cos(\omega t) \]

### Step 2: Calculate the effective (RMS) value of the AC component
The effective value (or RMS value) of an AC current \( I_{AC} = A \cos(\omega t) \) is given by:
\[ I_{AC, \text{eff}} = \frac{A}{\sqrt{2}} \]
For our AC current \( I_{AC} = 40 \cos(\omega t) \):
\[ I_{AC, \text{eff}} = \frac{40}{\sqrt{2}} = 20\sqrt{2} \, \text{A} \]

### Step 3: Calculate the effective value of the total current
The effective value of the total current \( I_{\text{eff}} \) when combining DC and AC currents is given by:
\[ I_{\text{eff}} = \sqrt{I_{DC}^2 + I_{AC, \text{eff}}^2} \]
Substituting the values:
\[ I_{\text{eff}} = \sqrt{10^2 + (20\sqrt{2})^2} \]

### Step 4: Simplify the expression
Calculating the squares:
\[ 10^2 = 100 \]
\[ (20\sqrt{2})^2 = 400 \times 2 = 800 \]
Now substitute back into the equation:
\[ I_{\text{eff}} = \sqrt{100 + 800} = \sqrt{900} \]

### Step 5: Calculate the final result
Taking the square root:
\[ I_{\text{eff}} = 30 \, \text{A} \]

Thus, the effective value of the resulting current is:
\[ \boxed{30 \, \text{A}} \]