To solve the problem, we need to find the reading of the AC ammeter when both DC and AC currents are flowing simultaneously in the circuit. Here’s a step-by-step solution:
### Step 1: Understand the Given Information
- When a direct current (DC) flows through the circuit, the AC ammeter reads **3 A**.
- When an alternating current (AC) flows through the circuit, the AC ammeter reads **4 A**.
### Step 2: Identify the RMS Values
- The reading of the ammeter for DC is equal to the actual current since DC is constant.
- Therefore, \( I_{DC} = 3 \, \text{A} \) (this is also the RMS value for DC).
- The reading of the ammeter for AC is given as 4 A, which is the RMS value of the AC current.
- Therefore, \( I_{AC, RMS} = 4 \, \text{A} \).
### Step 3: Find the Peak Value of AC Current
- The relationship between the RMS value and the peak value for AC is given by:
\[
I_{AC, peak} = I_{AC, RMS} \times \sqrt{2} = 4 \, \text{A} \times \sqrt{2} = 4\sqrt{2} \, \text{A}
\]
### Step 4: Combine DC and AC Currents
- When both currents are flowing simultaneously, the total instantaneous current can be expressed as:
\[
I(t) = I_{DC} + I_{AC}(t) = 3 + 4\sqrt{2} \sin(\omega t)
\]
### Step 5: Calculate the RMS Value of the Combined Current
- To find the RMS value of the total current, we need to square the total current expression and integrate over one complete cycle.
1. **Square the total current:**
\[
I^2(t) = (3 + 4\sqrt{2} \sin(\omega t))^2 = 9 + 24\sqrt{2} \sin(\omega t) + 32 \sin^2(\omega t)
\]
2. **Use the identity for \(\sin^2(\omega t)\):**
\[
\sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2}
\]
Thus,
\[
I^2(t) = 9 + 24\sqrt{2} \sin(\omega t) + 16(1 - \cos(2\omega t)) = 9 + 16 + 24\sqrt{2} \sin(\omega t) - 16\cos(2\omega t)
\]
\[
I^2(t) = 25 + 24\sqrt{2} \sin(\omega t) - 16\cos(2\omega t)
\]
3. **Integrate over one period \(T\):**
- The average value of \(\sin(\omega t)\) over one period is 0, and the average value of \(\cos(2\omega t)\) over one period is also 0.
- Therefore, the integral simplifies to:
\[
\text{Average of } I^2 = \frac{1}{T} \int_0^T I^2(t) dt = 25
\]
### Step 6: Calculate the RMS Value
- The RMS value is the square root of the average of \(I^2\):
\[
I_{RMS} = \sqrt{25} = 5 \, \text{A}
\]
### Final Answer
The reading of the AC ammeter when both DC and AC currents flow simultaneously is **5 A**.
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