In the circuit given below- the cells E1 and E2 have e-m-fs 4V and 8V and internal resistance 0-5- and 1- respectively- The potential difference across E1 and E2 will be respectively

The correct option is C 4.25V and 7.5V R_eff = r_1+r_2 + 4.5 +3 × 69 #160; #160; #160; #160; #160; = 1.5 +4.5 + 2 = 8 Ω I = ...

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Standard XII

Physics

Conductivity and Resistivity

In the circui...

Question

In the circuit given below, the cells $E_{1}$ and $E_{2}$ have e.m.fs $4 V$ and $8 V$ and internal resistance $0.5 Ω$ and $1 Ω$ respectively. The potential difference across $E_{1}$ and $E_{2}$ will be respectively

4 V

and

8 V

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3.75 V

and

8.5 V

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4.25 V

and

7.5 V

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4.25 V

and

8.5 V

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Solution

The correct option is C $4.25 V$ and $7.5 V$
$R_{e f f} = r_{1} + r_{2} + 4.5 + \frac{3 \times 6}{9}$
$= 1.5 + 4.5 + 2 = 8 Ω$
$I = \frac{E_{2} - E_{1}}{8} = 0.5 A$
$P . D a c r o s s E_{1} = E_{1} + I r_{1}$
$= 4 + 0.5 + 0.5 = 4.25 V$
$E_{2} = E_{2} - I r_{2} = 8 - 0.5 = 7.5 V$

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In the circuit given below, the cells E1 and E2 have e.m.fs 4V and 8V and internal resistance 0.5Ω and 1Ω respectively. The potential difference across E1 and E2 will be respectively

The correct option is C 4.25V and 7.5VReff=r1+r2+4.5+3×69=1.5+4.5+2=8ΩI=E2−E18=0.5AP.DacrossE1=E1+Ir1 =4+0.5+0.5=4.25VE2=E2−Ir2=8−0.5=7.5V

In the circuit given below, the cells $E_{1}$ and $E_{2}$ have e.m.fs $4 V$ and $8 V$ and internal resistance $0.5 Ω$ and $1 Ω$ respectively. The potential difference across $E_{1}$ and $E_{2}$ will be respectively

The correct option is C $4.25 V$ and $7.5 V$
$R_{e f f} = r_{1} + r_{2} + 4.5 + \frac{3 \times 6}{9}$
$= 1.5 + 4.5 + 2 = 8 Ω$
$I = \frac{E_{2} - E_{1}}{8} = 0.5 A$
$P . D a c r o s s E_{1} = E_{1} + I r_{1}$
$= 4 + 0.5 + 0.5 = 4.25 V$
$E_{2} = E_{2} - I r_{2} = 8 - 0.5 = 7.5 V$