Question

If the internal resistance of the cell is negligible, then the current flowing through the circuit is

• A. 0.04 A
• B. 0.06 A
• C. 0.08 A
• D. 0.10 A

Hint:

When analyzing a series circuit, the total resistance is simply the sum of individual resistances. The current is found using Ohm's law or Kirchhoff’s Voltage Law.

Answer 1

The Correct Option is B

Step 1: Understanding the circuit configuration.
In this problem, we are given a circuit with two diodes \( D_1 \) and \( D_2 \), two resistors of \( 30 \, \Omega \) each, and a \( 20 \, \Omega \) resistor connected in series with a \( 5 \, \text{V} \) battery. The internal resistance of the cell is negligible, which simplifies the analysis.
The diodes are assumed to be ideal, and the resistors \( 30 \, \Omega \) and \( 20 \, \Omega \) are in series with the battery and diodes.

Step 2: Apply Kirchhoff's Voltage Law (KVL).
To find the current in the circuit, we will apply Kirchhoff's Voltage Law (KVL), which states that the sum of all the voltages around a closed loop must be zero. In this circuit, the KVL equation is:
\[ V_{\text{battery}} - I \cdot R_1 - I \cdot R_2 - I \cdot R_3 = 0, \]
where:
- \( V_{\text{battery}} = 5 \, \text{V} \),
- \( R_1 = R_2 = 30 \, \Omega \) (the resistors connected in series),
- \( R_3 = 20 \, \Omega \) (the resistor in the lower part of the circuit),
- \( I \) is the current flowing through the circuit.

Step 3: Solving for the current \( I \).
We can now substitute the values into the KVL equation:
\[ 5 - I \cdot (30 + 30 + 20) = 0. \]
Simplifying: \[ 5 - I \cdot 80 = 0. \]
Solving for \( I \):
\[ I = \frac{5}{80} = 0.0625 \, \text{A}. \]

Step 4: Conclusion.
The current flowing through the circuit is approximately 0.06 A.
Final Answer:
Thus, the current flowing through the circuit is:
\[ \boxed{0.06 \, \text{A}}. \]