If in the circuit shown below, the internal resistance of the battery is 1.5 Ω and VP and VQ are the potentials at P and Q respectively, then what is the potential difference between the points P and Q ?

Question

If in the circuit shown below, the internal resistance of the battery is 1.5 Ω and VP and VQ are the potentials at P and Q respectively, then what is the potential difference between the points P and Q ? <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/66b634106910cf2b59235a0b483c779a-.png" /> (A) Zero (B) 4 V (VP > VQ) (C) 4 V (VQ > VP) (D) 2.5 V (VQ > VP)

Tardigrade Admin · Accepted Answer

R eq =(5/2) Ω i=(20/(5)2+1.5)=5 A <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/c5b761221b6ddc6f52b7d820979ffe08-.png" /> Potential difference between X and P, VX-VP=((5/2)) × 3=7.5 V ldots(i) VX-VQ=(5/2) × 2=5 V ldots (ii) On solving (i) and (ii), VP-VQ=-2.5 V, VQ>VP.

Q. If in the circuit shown below, the internal resistance of the battery is $1.5 Ω$ and $V_{P}$ and $V_{Q}$ are the potentials at $P$ and $Q$ respectively, then what is the potential difference between the points $P$ and $Q ?$

Zero

$4 V (V_{P} > V_{Q})$

$4 V (V_{Q} > V_{P})$

$2.5 V (V_{Q} > V_{P})$

$R_{e q} = \frac{5}{2} Ω$
$i = \frac{20}{\frac{5}{2} + 1.5} = 5 A$

Potential difference between $X$ and $P$ ,
$V_{X} - V_{P} = (\frac{5}{2}) \times 3 = 7.5 V \dots$ (i)
$V_{X} - V_{Q} = \frac{5}{2} \times 2 = 5 V \dots$ (ii)
On solving (i) and (ii),
$V_{P} - V_{Q} = - 2.5 V, V_{Q} > V_{P}$ .

Q. If in the circuit shown below, the internal resistance of the battery is 1.5Ω and VP​ and VQ​ are the potentials at P and Q respectively, then what is the potential difference between the points P and Q?

Zero

4V(VP​>VQ​)

4V(VQ​>VP​)

2.5V(VQ​>VP​)

Req​=25​Ω i=25​+1.520​=5A Potential difference between X and P, VX​−VP​=(25​)×3=7.5V…(i) VX​−VQ​=25​×2=5V… (ii) On solving (i) and (ii), VP​−VQ​=−2.5V,VQ​>VP​.

Q. If in the circuit shown below, the internal resistance of the battery is $1.5 Ω$ and $V_{P}$ and $V_{Q}$ are the potentials at $P$ and $Q$ respectively, then what is the potential difference between the points $P$ and $Q ?$

$4 V (V_{P} > V_{Q})$

$4 V (V_{Q} > V_{P})$

$2.5 V (V_{Q} > V_{P})$

$R_{e q} = \frac{5}{2} Ω$
$i = \frac{20}{\frac{5}{2} + 1.5} = 5 A$

Potential difference between $X$ and $P$ ,
$V_{X} - V_{P} = (\frac{5}{2}) \times 3 = 7.5 V \dots$ (i)
$V_{X} - V_{Q} = \frac{5}{2} \times 2 = 5 V \dots$ (ii)
On solving (i) and (ii),
$V_{P} - V_{Q} = - 2.5 V, V_{Q} > V_{P}$ .