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Question: A circuit consists of three batteries of emf \[{E_1} = 1{\text{V}}\], \({E_1} = 2{\text{V}}\) and \(...

A circuit consists of three batteries of emf E1=1V{E_1} = 1{\text{V}}, E1=2V{E_1} = 2{\text{V}} and E3=3V{E_3} = 3{\text{V}} and internal resistances 1Ω{\text{1}}\Omega , 2Ω2\Omega and 1Ω{\text{1}}\Omega respectively which are connected in parallel as shown in the figure. The potential difference between point P and Q is

A. 1.0V1.0{\text{V}}
B. 2.0V2.0{\text{V}}
C. 2.2V2.2{\text{V}}
D. 3.0V3.0{\text{V}}

Explanation

Solution

To solve this question, we need to use the formula for the equivalent emf of the batteries connected in parallel combination with each other. The batteries are connected in parallel combination across the point P and Q. Therefore the potential difference between the points P and Q will be equal to the equivalent emf of the batteries.

Formula used: The formulae used for solving this question are given by
EReq=E1R1+E2R2+..........+EnRn\dfrac{E}{{{R_{eq}}}} = \dfrac{{{E_1}}}{{{R_1}}} + \dfrac{{{E_2}}}{{{R_2}}} + .......... + \dfrac{{{E_n}}}{{{R_n}}}
1Req=1R1+1R2+..........+1Rn\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + .......... + \dfrac{1}{{{R_n}}}
Here EE is the equivalent emf of the parallel combination of the batteries of emfs E1{E_1}, E2{E_2},…., En{E_n} and internal resistances of R1{R_1}, R2{R_2},….., Rn{R_n} respectively. Also, Req{R_{eq}} is the equivalent internal resistance.

Complete step by step answer:
We know that the equivalent emf of a combination of batteries connected in parallel to each other is given by
EReq=E1R1+E2R2+..........+EnRn\dfrac{E}{{{R_{eq}}}} = \dfrac{{{E_1}}}{{{R_1}}} + \dfrac{{{E_2}}}{{{R_2}}} + .......... + \dfrac{{{E_n}}}{{{R_n}}} (1)
According to the question, we have E1=1V{E_1} = 1{\text{V}}, E2=2V{E_2} = 2{\text{V}}, and E3=3V{E_3} = 3{\text{V}}. Also, the internal resistances of these batteries are R1=1Ω{R_1} = {\text{1}}\Omega , R2=2Ω{R_2} = 2\Omega , and R3=1Ω{R_3} = 1\Omega . Substituting these in (1) we get
EReq=11+22+31\dfrac{E}{{{R_{eq}}}} = \dfrac{1}{1} + \dfrac{2}{2} + \dfrac{3}{1}
EReq=5\Rightarrow \dfrac{E}{{{R_{eq}}}} = 5 (2)
Also, since the internal resistances are connected in parallel combination with each other, the equivalent internal resistance of the combination is given by
1Req=1R1+1R2+..........+1Rn\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + .......... + \dfrac{1}{{{R_n}}}
Substituting R1=1Ω{R_1} = {\text{1}}\Omega , R2=2Ω{R_2} = 2\Omega , and R3=1Ω{R_3} = 1\Omega in the above equation, we get
1Req=11+12+11\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{1}
1Req=52\Rightarrow \dfrac{1}{{{R_{eq}}}} = \dfrac{5}{2}
Taking the reciprocal, we get
Req=25Ω{R_{eq}} = \dfrac{2}{5}\Omega
Substituting this in (2) we get
E2/5=5\dfrac{E}{{2/5}} = 5
E=25×5=2V\Rightarrow E = \dfrac{2}{5} \times 5 = 2{\text{V}}
So the equivalent emf of the given parallel combination of the batteries is equal to 2.0V2.0{\text{V}}. This means that the potential difference between the points P and Q is equal to 2.0V2.0{\text{V}}.
Hence, the correct answer is option B.

Note: There must be some drop across the equivalent internal resistance of the equivalent battery. This is because the circuit between points P and Q is not complete. So no current will flow through the equivalent battery, which means that the drop across the equivalent internal resistor will be equal to zero. Due to this we could take the potential difference is equal to the equivalent emf.