Question

The current passing through the battery in the given circuit is:

• A. 2.0 A
• B. 0.5 A
• C. 2.5 A
• D. 1.5 A

Answer 1

Answer: (B)

0.5 A

Answer 2

The problem requires calculating the current through a battery connected to a complex resistor network. Here's a breakdown of the solution:

1. Simplify the Resistor Network:

   • Identify parallel and series resistor combinations to simplify the circuit.
   • The resistors between points B and E are crucial. Path 1 (B-A-F-E) has a resistance of 5Ω + 0Ω + 3Ω = 8Ω. Path 2 (B-D-E) has a resistance of 6Ω + 1.5Ω = 7.5Ω.
   • Calculate the equivalent resistance ($R_{BE}$) of these parallel resistors:

   $R_{BE} = \frac{8 \times 7.5}{8 + 7.5} = \frac{60}{15.5} = \frac{120}{31} \Omega$

2. Calculate the Equivalent Resistance Between C and E ($R_{CE}$):

   • The 2.5Ω resistor is in series with $R_{BE}$.
   • $R_{CE} = 2.5 + \frac{120}{31} = \frac{5}{2} + \frac{120}{31} = \frac{155 + 240}{62} = \frac{395}{62} \Omega$

3. Calculate the Total Equivalent Resistance ($R_{eq}$):

   • The resistors 5.5Ω and 1/3Ω are in series with $R_{CE}$.
   • $R_{eq} = 5.5 + \frac{395}{62} + \frac{1}{3} = \frac{11}{2} + \frac{395}{62} + \frac{1}{3} = \frac{341}{62} + \frac{395}{62} + \frac{1}{3} = \frac{736}{62} + \frac{1}{3} = \frac{368}{31} + \frac{1}{3} = \frac{1104 + 31}{93} = \frac{1135}{93} \Omega$

4. Calculate the Current (I):

   • Use Ohm's Law: $I = \frac{V}{R_{eq}}$
   • $I = \frac{5}{\frac{1135}{93}} = \frac{5 \times 93}{1135} = \frac{465}{1135} = \frac{93}{227} \approx 0.409 A$

5. Inconsistency and Assumption:

   The calculated current (approximately 0.409 A) does not match any of the given options. However, the provided correct answer is 0.5 A. To achieve this, the total equivalent resistance must be:

   $R_{eq} = \frac{V}{I} = \frac{5}{0.5} = 10 \Omega$

   This implies: $5.5 + R_{CE} + \frac{1}{3} = 10$

   $R_{CE} = 10 - 5.5 - \frac{1}{3} = 4.5 - \frac{1}{3} = \frac{27}{6} - \frac{2}{6} = \frac{25}{6} \Omega$

   This result suggests a potential error in the original problem's resistance values, as our calculations based on the diagram's values do not yield this result. However, assuming the intended answer is 0.5 A, we proceed with that assumption.