Question

In the circuit shown, value of $I$ in ampere is

A. \[1\]
B. \[0.60\]
C. \[0.4\]
D. \[1.5\]

Answer 1

Hint: The given circuit has a combination of resistors both series and parallel, to find I, simplify the resistors to $R_{net}$ and use Ohm’s law.

Resistors in series: $R_{S}=R_{1}+R_{2}$
Resistors in parallel: $\dfrac{1}{R_{P}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}$
Ohm’s Law that: $V=IR$

Formula used:
Resistors in series: $R_{S}=R_{1}+R_{2}$
Resistors in parallel: $\dfrac{1}{R_{P}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}$
Ohm’s Law that: $V=IR$

Complete step-by-step answer:
To find the value of $I$ we need to simplify the diagram. This can be done by using the resistor in series and parallel formula.
Resistors in series: $R_{S}=R_{1}+R_{2}$ and resistors in parallel: $\dfrac{1}{R_{P}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}$
We also know from Ohm’s Law that: $V=IR$
Using the above, since two $4\Omega $ resistors are in parallel

$\dfrac{1}{R_{P}}=\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}$
$R_{P}=2\Omega$
Again, $2\Omega \; , 4\Omega $ resistors are in series

$R_{S}=4+2=6\Omega$
Again,$4\Omega \; , 6\Omega $ are in parallel

$ \dfrac{1}{R_{P}}=\dfrac{1}{4}+\dfrac{1}{6}=\dfrac{6+4}{6\times 4}=\dfrac{10}{2 4}$
$R_{P}=2.4\Omega$
Again, $2.4\Omega \; , 1.6\Omega $ resistors are in series

$R_{net}=2.4 +1.6=4\Omega$
$R_{net}=4\Omega$
Since $V=4V$
From Ohm’s law; $V=IR_{net}$
$4=I\times 4$
$I=1A$
Hence the answer is A. 1

Additional Information:
Resistors are said to be in series whenever the current flows through the resistor equally. The equivalent resistance of a series resistor is equal to the algebraic sum of the individual resistances. The current in the circuit depends on the voltage supplied by the voltage source and the potential crop across the resistor. As energy is conserved, the sum of potential drop due to each resistor is equal to the voltage source.
$V=V_{1}+V_{2}=IR_{1}+IR_{2}=I(R_{1}+R_{2})=IR_{net}$
$I=\dfrac{V}{R_{net}}$
$R_{net}=R_{1}+R_{2}$
Similarly, resistors are said to be in parallel when one end of the resistors are connected by a wire, and the other end by another wire, thus the potential drop across each resistor is the same. At the junction since different currents flow through the different resistors, the sum of the individual currents is equal to the current in the circuit.
$I=I_{1}+I_{2}=\dfrac{V}{R_{1}}+\dfrac{V}{R_{2}}$
$I=V(\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}})=\dfrac{V}{R_{net}}$
$\dfrac{1}{R_{net}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}$

Note: Be careful in identifying if the resistance is connected in series or parallel, and use the proper formula. While using resistors in parallel formula, don’t forget to take the reciprocals of the resistance. Also remember the principle behind the connections.