Answer

Verified

18k+ views

**Hint:**The question can be solved by grounding one end of the wire and then applying Kirchhoff’s Current Law to the loop formed. Kirchhoff’s Voltage Law can also be used to solve this problem.

**Complete Step by Step Solution:**Kirchhoff’s Current Law, deals with the conservation of charge entering and leaving a junction. Kirchhoff’s Current Law, often shortened to KCL, states that “The algebraic sum of all currents entering and exiting a node must equal zero.”

This law is used to describe how a charge enters and leaves a wire junction point or node on a wire.

Connect the lower wire of the $1\Omega $ resistor to the ground (Q end) and apply KCL. Consider the grounded circuit as shown below.

Applying, KCL at the Q point, we can write,

$\dfrac{{V + 6}}{3} + \dfrac{V}{1} = \dfrac{{9 - V}}{5}$

$ \Rightarrow V\left[ {\dfrac{1}{3} + \dfrac{1}{5} + 1} \right] = \dfrac{9}{5} - 2$

Simplifying the equation further,

$ \Rightarrow V\left[ {\dfrac{{5 + 3 + 15}}{{15}}} \right] = \dfrac{{9 - 10}}{5}$

The potential difference between points Q and P is given by,

$ \Rightarrow V = - \dfrac{1}{5} \times \dfrac{{15}}{{23}} = \dfrac{{ - 3}}{{23}} = - 0.13V$

Thus the current in the 1Ω resistor is ${\text{I = }}\dfrac{{\text{V}}}{{\text{R}}}{\text{ = }}\dfrac{{{\text{0}}{\text{.13}}}}{{\text{1}}}{\text{ = 0}}{\text{.13A}}$.

**The current flows from Q to P. The correct answer is Option C.**

**Note:**Kirchhoff's voltage law (KVL) states that the sum of all voltages around any closed loop in a circuit must be equal to zero. This is a consequence of charge conservation and also conservation of energy. This means that the sum of all potential differences across the component involved in the circuit gives a zero reading, as expected.

Assuming potential at Q is $V$, we apply KVL loop 1,

$9 - 2i - 1\left( {i - {i_1}} \right) - 3i = 0$.

When we apply KVL to loop 2,

$6 - 3{i_1} + 1\left( {i - {i_1}} \right) = 0$.

Solving the equations for the two loops,

${\text{i = 1}}{\text{.82A}}$ and ${{\text{i}}_{\text{1}}}{\text{ = 1}}{\text{.95A}}$.

Current through the $1\Omega $ resistor is ${\text{0}}{\text{.13A}}$.

Recently Updated Pages

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

Let f be a twice differentiable such that fleft x rightfleft class 11 maths JEE_Main

Find the points of intersection of the tangents at class 11 maths JEE_Main

For the two circles x2+y216 and x2+y22y0 there isare class 11 maths JEE_Main

The path difference between two waves for constructive class 11 physics JEE_MAIN

What is the difference between solvation and hydra class 11 chemistry JEE_Main

Other Pages

Derive an expression for maximum speed of a car on class 11 physics JEE_Main

Assertion Work and heat are two equivalent from of class 11 chemistry JEE_Main

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

Formal charge on nitrogen and oxygen in NO3 ion are class 11 chemistry JEE_Main

Why does capacitor block DC and allow AC class 12 physics JEE_Main

Electric field due to uniformly charged sphere class 12 physics JEE_Main