For the circuit shown, with R1 = 1.0 Ω , R2 = 2.0 Ω , E1 = 2 V and E2 = E3 = 4 V, the potential difference between the points 'a' and 'b' is approximately (in V) :

Question

For the circuit shown, with R1 = 1.0 Ω , R2 = 2.0 Ω , E1 = 2 V and E2 = E3 = 4 V, the potential difference between the points 'a' and 'b' is approximately (in V) : (A) 2.7 (B) 3.3 (C) 2.3 (D) 3.7

Tardigrade Admin · Accepted Answer

Eeq = ((E1/2R1) + (E2/R2) + (E3/2R1)/(1)2R1 + (1)R2 + (1/2R1) = ((2/2) + (4/2) + (4/2)/(1)2 + (1)2 + (1/2) = (5/(3)2) = (10/3) = 3.3

Q. For the circuit shown, with $R_{1} = 1.0Ω, R_{2} = 2.0Ω, E_{1} = 2 V$ and $E_{2} = E_{3} = 4 V$ , the potential difference between the points $^{'} a^{'}$ and $^{'} b^{'}$ is approximately (in $V$ ) :

2.7

3.3

2.3

3.7

$E_{e q} = \frac{\frac{E _{1}}{2 R _{1}} + \frac{E _{2}}{R _{2}} + \frac{E _{3}}{2 R _{1}}}{\frac{1}{2 R _{1}} + \frac{1}{R _{2}} + \frac{1}{2 R _{1}}}$
$= \frac{\frac{2}{2} + \frac{4}{2} + \frac{4}{2}}{\frac{1}{2} + \frac{1}{2} + \frac{1}{2}}$
$= \frac{5}{\frac{3}{2}} = \frac{10}{3} = 3.3$

Q. For the circuit shown, with R1​=1.0Ω,R2​=2.0Ω,E1​=2V and E2​=E3​=4V, the potential difference between the points ′a′ and ′b′ is approximately (in V) :

2.7

3.3

2.3

3.7

Eeq​=2R1​1​+R2​1​+2R1​1​2R1​E1​​+R2​E2​​+2R1​E3​​​ =21​+21​+21​22​+24​+24​​ =23​5​=310​=3.3

Q. For the circuit shown, with $R_{1} = 1.0Ω, R_{2} = 2.0Ω, E_{1} = 2 V$ and $E_{2} = E_{3} = 4 V$ , the potential difference between the points $^{'} a^{'}$ and $^{'} b^{'}$ is approximately (in $V$ ) :

$E_{e q} = \frac{\frac{E _{1}}{2 R _{1}} + \frac{E _{2}}{R _{2}} + \frac{E _{3}}{2 R _{1}}}{\frac{1}{2 R _{1}} + \frac{1}{R _{2}} + \frac{1}{2 R _{1}}}$
$= \frac{\frac{2}{2} + \frac{4}{2} + \frac{4}{2}}{\frac{1}{2} + \frac{1}{2} + \frac{1}{2}}$
$= \frac{5}{\frac{3}{2}} = \frac{10}{3} = 3.3$