
The circuit involves resistors in series and parallel. First, the two 4 \(\Omega\) resistors are in parallel. The total resistance of this parallel combination is: \[ \frac{1}{R_{\text{parallel}}} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \] \[ R_{\text{parallel}} = 2 \, \Omega \] Now, this 2 \(\Omega\) combination is in series with the 5 \(\Omega\) resistor, so the total resistance of this combination is: \[ R_{\text{total1}} = 2 + 5 = 7 \, \Omega \] Next, this 7 \(\Omega\) is in parallel with the 10 \(\Omega\) resistor: \[ \frac{1}{R_{\text{total}}} = \frac{1}{7} + \frac{1}{10} \] \[ \frac{1}{R_{\text{total}}} = \frac{10 + 7}{70} = \frac{17}{70} \] \[ R_{\text{total}} = \frac{70}{17} \approx 4.12 \, \Omega \] So the equivalent resistance between A and B is approximately \(4.12 \, \Omega\).
Now, using Ohm's law, the total current in the circuit is: \[ I = \frac{V}{R_{\text{total}}} = \frac{6}{4.12} \approx 1.46 \, \text{A} \] The potential difference across the 5 \(\Omega\) resistor can be found using: \[ V = I \times 5 = 1.46 \times 5 = 7.3 \, \text{V} \]
In the circuit given below, five resistances of \(10 \Omega\), \(40 \Omega\), \(20 \Omega\), \(30 \Omega\), and \(60 \Omega\) are connected as shown to a battery of 18 volts.
Potential difference between points A and B
In the circuit given below, five resistances of \(10 \Omega\), \(40 \Omega\), \(20 \Omega\), \(30 \Omega\), and \(60 \Omega\) are connected as shown to a battery of 18 volts.
Total resistance of the circuit
In the circuit given below, five resistances of \(10 \Omega\), \(40 \Omega\), \(20 \Omega\), \(30 \Omega\), and \(60 \Omega\) are connected as shown to a battery of 18 volts.
Total current flowing in the circuit
Symbols of few components used in electrical circuits are given in column A. Match with names given in column B and choose the correct option:
In the given circuit, find:
(i) Total resistance
(ii) Current through the cell, and
(iii) Potential difference across 0.8 \(\Omega\) resistance. 