Solveeit Logo

Question

Question: Determine the potential drop between points \(A\) and \(C\) in the following circuit. Resistance \[1...

Determine the potential drop between points AA and CC in the following circuit. Resistance 1  Ω1\;\Omega and 2Ω2\Omega ( internal resistance of cell)
(A).45V(A). \dfrac{4}{5}V
(B).1.75V(B). 1.75V
(C).2.25 V(C). 2.25{\text{ }}V
(D).54V(D). \dfrac{5}{4}V

Explanation

Solution

First, the combination of the resistances given here has to be observed and to find the equivalent resistance of the circuit.
Second, the current that flows through the circuit has to be calculated by using the given EMF and equivalent resistance of the circuit.
Finally by using the calculated current find the potential drop of the required terminals.

Formula used:
The equivalent resistance R=R1+R2+R3R = {R_1} + {R_2} + {R_3} [ since they are in series connection]
Current flowing through the circuit, I=E2E1RI = \dfrac{{{E_2} - {E_1}}}{R}
E2{E_2} and E1{E_1} are two EMFs given in the circuit.
The voltage drop between the two points AAand CC, VAC=E1+IR1{V_{AC}} = {E_1} + I{R_1}

Complete step by step answer:
The circuit given below is showing that two batteries are connected between terminals CC and AA is 2Volt2Volt, and terminals AA and BB is 4Volt4Volt. There are two resistances connected along with two batteries and one battery is in series with them.

Let us consider, R1=1Ω,R2=2Ω{R_1} = 1\Omega ,{R_2} = 2\Omega and R3=5Ω{R_3} = 5\Omega .
And, V1=2V,V2=4V{V_1} = 2V,{V_2} = 4V
The two EMFs are in opposite directions.

The two EMFs are in opposite directions.
since, E2>E1{E_2} > {E_1} , the current will flow from point BB to CC.
Hence, The amount of the current flowing through the circuit, I=V2V1RI = \dfrac{{{V_2} - {V_1}}}{R}
Where, RR is the equivalent resistance of the circuit.
R=R1+R2+R3R = {R_1} + {R_2} + {R_3}
R=(1+2+5)=8Ω\Rightarrow R = (1 + 2 + 5) = 8\Omega
So, I=V2V1RI = \dfrac{{{V_2} - {V_1}}}{R}
I=428\Rightarrow I = \dfrac{{4 - 2}}{8}
I=14\Rightarrow I = \dfrac{1}{4}
I=428=0.25A\Rightarrow I = \dfrac{{4 - 2}}{8} = 0.25A
The voltage drop between the two points AA and CC, VAC=E1+IR1{V_{AC}} = {E_1} + I{R_1}
VAC=2+(0.25×1)\Rightarrow {V_{AC}} = 2 + (0.25 \times 1)
VAC=2.25Volt\Rightarrow {V_{AC}} = 2.25Volt
So the potential drop between the two points AA and CC is 2.25Volt2.25Volt.

Hence, the correct answer is option (C).

Note:
• If there is no resistance or any source voltage between two points of a circuit, the potential will be the same for two points.
• The EMF can be defined as the difference between the potentials of the two terminals. We can say, the potential of the negative terminal is greater than the positive terminal how much the EMF is.