
In an \[LCR\] series a.c. circuit, the voltage across each of the components \[L,{\text{ }}C\] and \[R\] is \[50{\text{ }}V\]. The voltage across the \[LC\] combination will be:
A) $50 V$
B) $50\sqrt 2 V$
C) $100 V$
D) $0 V$ (zero)
Answer
525.6k+ views
Hint: In \[LCR\] series a.c circuit, the components \[L,{\text{ }}C\] and \[R\] are related to each other in a way that the voltage across the inductor \[L\](${V_L}$) and voltage across the capacitor \[C\](${V_C}$) has a constant phase difference. The voltage across the resistance \[R\](${V_R}$) and current($i$) does not have any phase difference.
Complete step by step solution:
According to the question, a \[LCR\] series a.c. circuit is given. The voltage across \[L,{\text{ }}C\] and \[R\] is \[50{\text{ }}V\].
We know that in an \[LCR\] series circuit, the voltage across the inductor \[L\](${V_L}$) leads the current($i$) by \[{90^ \circ }\] and voltage across the capacitor \[C\](${V_C}$) lags the current($i$) by \[{90^ \circ }\]. So, the inductance and the capacitance are in opposite phases. In an \[LCR\] series circuit, the voltage across the resistance \[R\](${V_R}$) is in the same phase with current($i$).
So, the voltage across the \[LC\] combination will be given as:
$
{V_{LC}} = {V_L} - {V_C} \\
\Rightarrow {V_{LC}} = 50 - 50 \\
\Rightarrow {V_{LC}} = 0 \\
$
To understand the phase difference in different voltages, we can make a graph which shows the phase difference between voltages.
According to the above graph, ${V_R}$ and $i$ are in the same phase. ${V_L}$leads current $i$ by \[{90^ \circ }\] and ${V_C}$ lags current $i$ by \[{90^ \circ }\]. So, the voltage across \[LC\] combination is zero.
Hence, option (D) is correct.
Note: Voltage across the inductor ${V_L}$ and current $i$ has a phase difference. Voltage across the capacitor ${V_C}$ and current $i$ has a phase difference. The voltage across the resistance ${V_R}$ and current $i$ has zero phase difference. According to the graph, current $i$ and voltage across the resistance ${V_R}$ are plotted on X-axis.
Complete step by step solution:
According to the question, a \[LCR\] series a.c. circuit is given. The voltage across \[L,{\text{ }}C\] and \[R\] is \[50{\text{ }}V\].

We know that in an \[LCR\] series circuit, the voltage across the inductor \[L\](${V_L}$) leads the current($i$) by \[{90^ \circ }\] and voltage across the capacitor \[C\](${V_C}$) lags the current($i$) by \[{90^ \circ }\]. So, the inductance and the capacitance are in opposite phases. In an \[LCR\] series circuit, the voltage across the resistance \[R\](${V_R}$) is in the same phase with current($i$).
So, the voltage across the \[LC\] combination will be given as:
$
{V_{LC}} = {V_L} - {V_C} \\
\Rightarrow {V_{LC}} = 50 - 50 \\
\Rightarrow {V_{LC}} = 0 \\
$
To understand the phase difference in different voltages, we can make a graph which shows the phase difference between voltages.

According to the above graph, ${V_R}$ and $i$ are in the same phase. ${V_L}$leads current $i$ by \[{90^ \circ }\] and ${V_C}$ lags current $i$ by \[{90^ \circ }\]. So, the voltage across \[LC\] combination is zero.
Hence, option (D) is correct.
Note: Voltage across the inductor ${V_L}$ and current $i$ has a phase difference. Voltage across the capacitor ${V_C}$ and current $i$ has a phase difference. The voltage across the resistance ${V_R}$ and current $i$ has zero phase difference. According to the graph, current $i$ and voltage across the resistance ${V_R}$ are plotted on X-axis.
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