Physics - Current Electricity Question with Solution | TestHub

PhysicsCurrent ElectricityCircuit AnalysisHard2 minQB

PhysicsHardsingle choice

Passage / Comprehension

In the circuit shown, both batteries are ideal. The electromotive force (EMF) $E$ of battery 1 has a fixed value, but the EMF $E$ of battery 2 can be varied between 1 V and 10 V. The graph gives the currents through the two batteries as a function of $E$ , but the plots are not marked as to which corresponds to which battery. For both plots, current is assumed to be negative when its direction through the battery is opposite the direction of that battery's EMF (i.e., from negative to positive).

The resistance $R_{1}$ has value

Options:

Answer:

Solution:

$I_{1} = \frac{E _{1} - E _{2}}{R _{1}}, I_{1} + I_{2} = \frac{E _{2}}{R _{2}} \Rightarrow I_{2} = \frac{E _{2}}{R _{2}} - \frac{E _{1} - E _{2}}{R _{1}}$ $\Rightarrow I_{1} = (\frac{- 1}{R _{1}}) E_{2} + \frac{E _{1}}{R _{1}} & I_{2} = (\frac{1}{R _{1}} + \frac{1}{R _{2}}) E_{2} - \frac{E _{1}}{R _{1}}$ $\frac{1}{R _{1}} = \frac{0.3}{6} \Rightarrow R_{1} = 20Ω$ and $\frac{1}{R _{1}} + \frac{1}{R _{2}} = \frac{0.3}{4} \Rightarrow R_{2} = 40Ω$

Now as $\frac{E _{1}}{R _{1}} = 0.3 \Rightarrow E_{1} = 0.3 \times 20 = 6 V$

Stream:JEESubject:PhysicsTopic:Current ElectricitySubtopic:Circuit Analysis

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...