Question

Add the following voltages in the form given
Va=100sin(wt-45°)
Vb=50sin(wt+30°)
Vc=60cos(wt)

Answer 1

To add the given voltages \( V_a = 100 \sin(\omega t - 45^\circ) \), \( V_b = 50 \sin(\omega t + 30^\circ) \), and \( V_c = 60 \cos(\omega t) \), we need to convert all terms into a common format, preferably into sine or cosine functions.

### Step 1: Convert \( V_c \) to sine form

Since \( V_c = 60 \cos(\omega t) \), we can convert it to a sine function:

\[
V_c = 60 \cos(\omega t) = 60 \sin\left(\omega t + 90^\circ\right)
\]

### Step 2: Rewrite each voltage

Now we can rewrite all the voltages in sine form:

- \( V_a = 100 \sin(\omega t - 45^\circ) \)
- \( V_b = 50 \sin(\omega t + 30^\circ) \)
- \( V_c = 60 \sin(\omega t + 90^\circ) \)

### Step 3: Use the sine addition formula

Using the sine addition formula, we can express \( V_a \), \( V_b \), and \( V_c \) in terms of their sine and cosine components:

#### \( V_a \)

\[
V_a = 100 \left( \sin(\omega t)\cos(45^\circ) - \cos(\omega t)\sin(45^\circ) \right) = 100 \left( \sin(\omega t) \frac{\sqrt{2}}{2} - \cos(\omega t) \frac{\sqrt{2}}{2} \right)
\]

\[
= 50\sqrt{2} \sin(\omega t) - 50\sqrt{2} \cos(\omega t)
\]

#### \( V_b \)

\[
V_b = 50 \left( \sin(\omega t)\cos(30^\circ) + \cos(\omega t)\sin(30^\circ) \right) = 50 \left( \sin(\omega t) \frac{\sqrt{3}}{2} + \cos(\omega t) \frac{1}{2} \right)
\]

\[
= 25\sqrt{3} \sin(\omega t) + 25 \cos(\omega t)
\]

#### \( V_c \)

\[
V_c = 60 \sin(\omega t + 90^\circ) = 60 \sin(\omega t)\cos(90^\circ) + 60 \cos(\omega t)\sin(90^\circ) = 60 \cos(\omega t)
\]

### Step 4: Combine all voltages

Now let's combine the components of \( \sin(\omega t) \) and \( \cos(\omega t) \):

**Sine terms**:
\[
V_{total, \sin} = (50\sqrt{2} + 25\sqrt{3}) \sin(\omega t)
\]

**Cosine terms**:
\[
V_{total, \cos} = (-50\sqrt{2} + 25) \cos(\omega t)
\]

### Step 5: Resulting voltage

Now we can write the resulting voltage as:

\[
V_{total} = (50\sqrt{2} + 25\sqrt{3}) \sin(\omega t) + (-50\sqrt{2} + 25) \cos(\omega t)
\]

This is the combined voltage in the sine/cosine form. If you want it in the standard sine form again, you can convert it using the resultant amplitude and phase angle.

### Conclusion

The overall voltage can be expressed as:

\[
V_{total} = A \sin(\omega t + \phi)
\]

where \( A = \sqrt{(50\sqrt{2} + 25\sqrt{3})^2 + (-50\sqrt{2} + 25)^2} \) and \( \phi = \tan^{-1}\left(\frac{-50\sqrt{2} + 25}{50\sqrt{2} + 25\sqrt{3}}\right) \).

You would compute these numeric values if needed for final analysis.