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NEET 2015
Exam
NEET
Year
2015
Subject
Physics
Chapter
Current Electricity

NEET 2015 · Physics · Current Electricity

A , B and C are voltmeters of resistance R , 1.5 R and 3 R...

A\text{A} , B\text{B} and C\text{C} are voltmeters of resistance R\text{R} , 1.5 R\text{1}\text{.5 R} and 3 R\text{3 R} respectively as shown in the figure. When some potential difference is applied between X\text{X} and Y\text{Y} , the voltmeter readings are VA,VB{V}_{A}, {V}_{B} and VC{V}_{C} respectively. Then:
A , B and C are voltmeters of resistance R , 1.5 R and 3 R respectively as shown in the figure. When some potential

Options

  1. AVA=VB=VC{V}_{A} = {V}_{B} = {V}_{C}
  2. BVA≠VB=VC{V}_{A} \ne {V}_{B} = {V}_{C}
  3. CVA=VB≠VC{V}_{A} = {V}_{B} \ne {V}_{C}
  4. DVA≠VB≠VC{V}_{A} \ne {V}_{B} \ne {V}_{C}

Correct answer

A. VA=VB=VC{V}_{A} = {V}_{B} = {V}_{C}

Step-by-step solution

A , B and C are voltmeters of resistance R , 1.5 R and 3 R respectively as shown in the figure. When some potential Voltage in parallel is constant iP=iB+iCVB=VcVpRi=VBRB+VCRC1Rp=1RB+1RC⇒1R′=23R+13R⇒R′=RVXP=VA=iRVPQ=VB=VC=iR⇒VA=VB=VC\begin{aligned} & i_P=i_B+i_C V_B=V_c \\ & \frac{V_p}{R_i}=\frac{V_B}{R_B}+\frac{V_C}{R_C} \\ & \frac{1}{R_p}=\frac{1}{R_B}+\frac{1}{R_C} \Rightarrow \frac{1}{R \prime}=\frac{2}{3 R}+\frac{1}{3 R} \\ \Rightarrow R^{\prime} & =R \\ & V_{X P}=V_A=i R \\ & V_{P Q}=V_B=V_C=i R \\ \Rightarrow V_A & =V_B=V_C \end{aligned})$
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