Question

Calculate the value of the current passing through the battery in the given circuit diagram.

Hint:

When multiple paths exist in a circuit, reduce parallel and series combinations step-by-step. Use Ohm’s law only after obtaining the total resistance.

Answer 1

From the circuit: - \( R_{AB} = 10\,\Omega \) - Between B and D, there are two paths: - Top path: \( R_{BD1} = 20\,\Omega \) - Bottom path: Series of: \[ R_{BC} = 5\,\Omega, \quad R_{CD1} = 40\,\Omega, \quad R_{CD2} = 20\,\Omega \text{ in parallel} \] Step 1: Calculate equivalent resistance of \( R_{CD1} \parallel R_{CD2} \) \[ \frac{1}{R_{CD}} = \frac{1}{40} + \frac{1}{20} = \frac{3}{40} \Rightarrow R_{CD} = \frac{40}{3} \, \Omega \] Step 2: Add with \( R_{BC} \) \[ R_{BCD} = 5 + \frac{40}{3} = \frac{55}{3} \, \Omega \] Step 3: Now calculate equivalent of \( R_{BD1} = 20 \, \Omega \) and \( R_{BCD} = \frac{55}{3} \, \Omega \) in parallel: \[ \frac{1}{R_{BD}} = \frac{1}{20} + \frac{3}{55} = \frac{11 + 12}{220} = \frac{23}{220} \Rightarrow R_{BD} = \frac{220}{23} \, \Omega \] Step 4: Add in series with \( R_{AB} = 10\,\Omega \) \[ R_{\text{total}} = 10 + \frac{220}{23} = \frac{450}{23} \, \Omega \] Step 5: Use Ohm's Law to find current: \[ I = \frac{V}{R} = \frac{6}{450/23} = \frac{138}{450} = \frac{23}{75} \, \text{A} \approx 0.307 \, \text{A} \]