To find the current in the circuit, we first need to calculate the total equivalent resistance (\(R_{eq}\)) of the resistors and then apply Ohm's Law.
Calculate Parallel Resistance
Two resistors, each with resistance \(R = 10\ \Omega\), are connected in parallel. The equivalent resistance (\(R_p\)) of parallel resistors is calculated using the formula:
\(R_p = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}}\)
Substituting the values:
\(R_p = \frac{1}{\frac{1}{10\ \Omega} + \frac{1}{10\ \Omega}} = \frac{1}{\frac{2}{10\ \Omega}} = \frac{10\ \Omega}{2} = 5\ \Omega\)
Calculate Total Equivalent Resistance
The parallel combination (\(R_p = 5\ \Omega\)) is connected in series with a third resistor (\(R_3 = 10\ \Omega\)). The total equivalent resistance (\(R_{eq}\)) for resistors in series is the sum of their resistances:
\(R_{eq} = R_p + R_3\)
Substituting the values:
\(R_{eq} = 5\ \Omega + 10\ \Omega = 15\ \Omega\)
Calculate Circuit Current
The total voltage (\(V\)) provided by the battery is 6 V. Using Ohm's Law, the current (\(I\)) in the circuit is given by:
\(I = \frac{V}{R_{eq}}\)
Substituting the values:
\(I = \frac{6\ V}{15\ \Omega} = 0.4\ A\)
Therefore, the current in the circuit is 0.4 A.
Similar Questions
- Two resistors, one of \(20\ \Omega\) and the other of \(30\ \Omega\), are connected in parallel. This combination is connected in series with an \(8\ \Omega\) resistor and a 12-V battery. The current through the \(20\ \Omega\) resistor is:
- Two resistors, each of \(20\ \Omega\), are connected in parallel, and this combination is connected across a 40-V supply. Find the voltage across each resistor.
- Two \(100\text{-}\Omega\) resistors are connected in parallel, and this combination is connected across a 40-V supply. Find the current supplied by the voltage source.
- Two resistors 1 ohm and 2 ohm are connected in series. The effective resistance is
- Two resistors, one of \(8\ \Omega\) and the other of \(24\ \Omega\) are connected in parallel. This combination is connected in series with a \(14\ \Omega\) resistor and a 12 V battery. The current in the \(8\ \Omega\) resistor is:
- Two resistors, A and B are connected in parallel combination in an electric circuit as to have an equivalent resistance of \(3\,\Omega\). On connecting the resistors A and B in series combination the equivalent resistant is found to be \(12\,\Omega\). Then the resistance R1 and R2 of the two resistors A and B respectively could be
- Two resistors, one of \(20\ \Omega\) and the other of \(30\ \Omega\), are connected in parallel. This combination is connected in series with an \(8\ \Omega\) resistor and a 12-V battery. The current through the \(20\ \Omega\) resistor is:
- Two resistors, each of \(20\ \Omega\), are connected in parallel, and this combination is connected across a 40-V supply. Find the voltage across each resistor.
- Two \(100\ \Omega\) resistors are connected in parallel, and this combination is connected across a 40-V supply. Find the current supplied by the voltage source.
Important Questions from Combination of Resistors — Series and Parallel
A cell of negligible resistance and e.m.f 2 volt is connected to series combination of 2 ohm, 3 ohm and 5 ohm. The potential difference across the 3 ohm resistance is:
3 resistors of 3 ohm each connected in series. What is the mean values of resistors?
Ten cells, each of 2 volts emf and 1 ohm internal resistance are connected in series. What is the current flowing through a resistance of 10 ohms connected across this combination of cells?
Three resistor, each equal to 3 Ω are connected so as to form a triangle. The equivalent resistance between any two vertices of the triangle is:
- Two resistors R 1 and R 2 arranged in parallel combination in an electrical closed circuit are made of the same material and of the same thickness. If the length of R 2 is twice the length of R 1, then the total resistance R satisfies
