Question

What is the potential drop between points A and C in the following circuit ? Resistances $1 \, \Omega $ and $2 \, \Omega $ represent the internal resistances of the respective cells.

• A. $1.75\,V$
• B. $2.25\,V$
• C. $ \frac{5}{4}{V} $
• D. $ \frac{4}{5} V $

Answer 1

The Correct Option is B

Emf's $E_{1}$ and $E_{2}$ are opposing each other.
Since, $E_{2}>E_{1}$ current will move from right to left.

Current in circuit
$i=\frac{E_{2}-E_{1}}{R+r_{1}+r_{2}}$
$=\frac{4-2}{5+1+2}=\frac{2}{8}=0.25 A$
The potential drop between points $A$ and $C$ is
$V_{A}-V_{C} =E_{1}+i r_{1} $
$=2+(0.25 \times 1) $
$=2.25 \,V$

Answer 2

The circuit given below is showing that two batteries are connected between terminals 

C and A are 2 Volt and terminals A and B are 4 Volt. There are two resistances connected along with two batteries and one battery is in series with them.

 

Let us consider, 

R₁=1Ω,R₂=2Ω and R₃=5Ω

And, V₁=2V,V₂=4V

The two EMFs are in opposite directions.

The two EMFs are in opposite directions.

since 

E₂>E₁, the current will flow from point B to C.

 

Hence, The amount of current flowing through the circuit, 

I=\(\frac{V_2-V_1}{R}\)

Where R is the equivalent resistance of the circuit.

R=R₁+R₂+R₃

⇒ R=(1+2+5)=8Ω

So, 

I=\(\frac{V_2 - V_1}{R}\)

⇒ I=4−\(\frac{2}{8}\)

 

⇒ I=\(\frac{1}{4}\)

⇒ I=4−\(\frac{2}{8}\)

= 0.25A

The voltage drop between the two points 

 

Answer 3

Here 2 Emf's E₁ and E₂ respectively oppose each other. Since,

E₂>E₁, so the current goes from right to left

Current in the circuit i =  emf or Total Resistance = \(\frac{E_2 - E_1}{R+r_1 + r_2}\)

​

​

 

Given, R=5Ω,r₁=1Ω,r₂=2Ω

E₁=2V and E₂=4V

∴ i =\(\frac{(4-2)}{(5+1+2)}\) = 0.25 A

The potential difference between 2 points (A, C) is,

V_A−V_C= E₁+ir₁

​

=2+0.25×1

=2.25 V

The correct option is (B).