Q132 / 180 · COMEDK 2022 · Physics
A DC ammeter and a hot wire ammeter are connected to a circuit in series. When a direct current is passed through circuit, the DC ammeter shows $6 \mathrm{~A}$. When $\mathrm{AC}$ current flows through circuit, what is the average readings in $\mathrm{DC}$ ammeter and the $\mathrm{AC}$ ammeter, if $\mathrm{DC}$ and AC currents flow simultaneously through the circuit?
- Correct answerA$\mathrm{DC}=6 \mathrm{~A}, \mathrm{AC}=10 \mathrm{~A} \quad$
- B$\mathrm{DC}=3 \mathrm{~A}, \mathrm{AC}=5 \mathrm{~A}$
- C$\mathrm{DC}=5 \mathrm{~A}, \mathrm{AC}=8 \mathrm{~A} \quad$
- D$\mathrm{DC}=2 \mathrm{~A}, \mathrm{AC}=3 \mathrm{~A}$
Solution
A DC ammeter measures the average value of the current passing through it. When a DC current $I_{DC}$ and an AC current $I_{AC}(t) = I_0 \sin(\omega t)$ flow simultaneously, the total current is $I(t) = I_{DC} + I_0 \sin(\omega t)$.
The reading of the DC ammeter is the average value of the total current over a full cycle of the AC component. Since the average value of a sinusoidal current over a full cycle is zero, the DC ammeter reads $I_{DC} = 6 \text{ A}$.
A hot wire ammeter measures the root mean square (RMS) value of the current. The total current is $I(t) = I_{DC} + I_0 \sin(\omega t)$. The RMS value $I_{rms}$ is given by $\sqrt{\dfrac{1}{T} \int_{0}^{T} (I_{DC} + I_0 \sin(\omega t))^2 dt}$.
Expanding the square: $I^2(t) = I_{DC}^2 + I_0^2 \sin^2(\omega t) + 2 I_{DC} I_0 \sin(\omega t)$.
Taking the average over a period $T$: $\langle I^2 \rangle = I_{DC}^2 + I_0^2 \langle \sin^2(\omega t) \rangle + 2 I_{DC} I_0 \langle \sin(\omega t) \rangle$.
Since $\langle \sin^2(\omega t) \rangle = \dfrac{1}{2}$ and $\langle \sin(\omega t) \rangle = 0$, we get $\langle I^2 \rangle = I_{DC}^2 + \dfrac{I_0^2}{2}$.
Thus, $I_{rms} = \sqrt{I_{DC}^2 + I_{AC, rms}^2}$. Given $I_{DC} = 6 \text{ A}$ and assuming the AC component is $8 \text{ A}$ (as per standard problem values for this specific question setup where $I_{rms} = \sqrt{6^2 + 8^2} = 10 \text{ A}$), the readings are $6 \text{ A}$ and $10 \text{ A}$.
Answer: $\mathrm{DC}=6 \mathrm{~A}, \mathrm{AC}=10 \mathrm{~A} \quad$
The reading of the DC ammeter is the average value of the total current over a full cycle of the AC component. Since the average value of a sinusoidal current over a full cycle is zero, the DC ammeter reads $I_{DC} = 6 \text{ A}$.
A hot wire ammeter measures the root mean square (RMS) value of the current. The total current is $I(t) = I_{DC} + I_0 \sin(\omega t)$. The RMS value $I_{rms}$ is given by $\sqrt{\dfrac{1}{T} \int_{0}^{T} (I_{DC} + I_0 \sin(\omega t))^2 dt}$.
Expanding the square: $I^2(t) = I_{DC}^2 + I_0^2 \sin^2(\omega t) + 2 I_{DC} I_0 \sin(\omega t)$.
Taking the average over a period $T$: $\langle I^2 \rangle = I_{DC}^2 + I_0^2 \langle \sin^2(\omega t) \rangle + 2 I_{DC} I_0 \langle \sin(\omega t) \rangle$.
Since $\langle \sin^2(\omega t) \rangle = \dfrac{1}{2}$ and $\langle \sin(\omega t) \rangle = 0$, we get $\langle I^2 \rangle = I_{DC}^2 + \dfrac{I_0^2}{2}$.
Thus, $I_{rms} = \sqrt{I_{DC}^2 + I_{AC, rms}^2}$. Given $I_{DC} = 6 \text{ A}$ and assuming the AC component is $8 \text{ A}$ (as per standard problem values for this specific question setup where $I_{rms} = \sqrt{6^2 + 8^2} = 10 \text{ A}$), the readings are $6 \text{ A}$ and $10 \text{ A}$.
Answer: $\mathrm{DC}=6 \mathrm{~A}, \mathrm{AC}=10 \mathrm{~A} \quad$
Hence, the correct answer is option: 'A'.