First slide

Kirchoff's Law

Question

Four resistance 40 $Ω$ , 60 $Ω$ , 90 $Ω$ and 110 $Ω$ make the arms of a quadrilateral ABCD. Across AC is a battery of emf 40 V and internal resistance negligible. The potential difference across BD is V is ______

Moderate

Solution

$\begin{array}{l} I n u p p e r b r a n c h, A B C A \\ + 40 - 40 i_{1} - 60 i_{1} = 0 \\ \Rightarrow 40 = 100 i_{1} \Rightarrow i_{1} = \frac{4}{10} A \\ I n l o w e r b r a n c h, A D C A \\ + 40 - 90 i_{2} - 110 i_{2} = 0 \\ \Rightarrow 40 = 200 i_{2} \\ \Rightarrow i_{2} = \frac{4}{20} A \\ L e t s a y p o t e n t i a l o f A = 40 V \\ \Rightarrow 40 - 40 i_{1} - V_{B} = 0 \\ \Rightarrow V_{B} = 40 - (40 \times \frac{4}{10}) \\ a n d \\ \Rightarrow 40 - 90 i_{2} - V_{D} = 0 \\ V_{D} = 40 - (90 \times \frac{4}{20}) \\ ∴ V_{D} - V_{B} = 40 \times \frac{4}{10} - 90 \times \frac{4}{20} = - 2 V \\ \Rightarrow V_{B} - V_{D} = + 2 V \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App