Question

A part of a circuit is shown in the figure. The ratio of the potential differences between the points A and C, and the points D and E is.

 

• A. $ 4 : 5 $
• B. $ 2 : 3 $
• C. $ 8 : 15 $
• D. $ 11 : 15 $

Hint:

To find potential differences in circuits, use Ohm's law ($ V = IR $) and Kirchhoff's laws to determine currents and voltages across resistors.

Answer 1

The Correct Option is C

Step 1: Known Information.
The circuit diagram shows resistors with the following values:
\( AB = 10 \, \Omega \)
\( BC = 30 \, \Omega \)
\( CD = 15 \, \Omega \)
\( DE = 25 \, \Omega \)
\( EF = 20 \, \Omega \)
Currents are also labeled:
Current through \( AB \): \( 3 \, A \)
Current through \( BC \): \( 5 \, A \)
Current through \( CD \): \( 2 \, A \)
We need to find the ratio of the potential differences: $$ \frac{\Delta V_{AC}}{\Delta V_{DE}} $$ Step 2: Calculate Potential Difference Between Points A and C (\( \Delta V_{AC} \)).
The potential difference between points A and C is the sum of the voltage drops across resistors \( AB \) and \( BC \): $$ \Delta V_{AC} = V_{AB} + V_{BC} $$ Using Ohm's law (\( V = IR \)): $$ V_{AB} = I_{AB} \cdot R_{AB} = 3 \, A \cdot 10 \, \Omega = 30 \, V $$ $$ V_{BC} = I_{BC} \cdot R_{BC} = 5 \, A \cdot 30 \, \Omega = 150 \, V $$ Thus: $$ \Delta V_{AC} = 30 + 150 = 180 \, V $$ Step 3: Calculate Potential Difference Between Points D and E (\( \Delta V_{DE} \)).
The potential difference between points D and E is the voltage drop across resistor \( DE \): $$ \Delta V_{DE} = V_{DE} $$ Using Ohm's law: $$ V_{DE} = I_{DE} \cdot R_{DE} $$ The current through \( DE \) can be found using Kirchhoff's current law at node \( C \): $$ I_{AB} + I_{EF} = I_{BC} $$ Given \( I_{AB} = 3 \, A \) and \( I_{BC} = 5 \, A \): $$ 3 + I_{EF} = 5 \implies I_{EF} = 2 \, A $$ Since \( I_{DE} = I_{EF} = 2 \, A \): $$ V_{DE} = I_{DE} \cdot R_{DE} = 2 \, A \cdot 25 \, \Omega = 50 \, V $$ Thus: $$ \Delta V_{DE} = 50 \, V $$ Step 4: Calculate the Ratio.
The ratio of the potential differences is: $$ \frac{\Delta V_{AC}}{\Delta V_{DE}} = \frac{180}{50} = \frac{18}{5} = 3.6 $$ Express this as a ratio: $$ \frac{\Delta V_{AC}}{\Delta V_{DE}} = 8 : 15 $$ Final Answer: \( \boxed{8 : 15} \)