Question

Find out total current supplied to the given circuit by the battery.​

Answer 1

Answer:

Total current supplied to the circuit by the battery is, I=ReqV=6Ω6V=1A.

Answer 2

The circuit can be rearranged as in fig. 1.

Here the section CDEF is a star connection which can be replaced by delta connection of resistors having resistance $\small\sf{R_1,\ R_2}$ and $\small\sf{R_3}$ each as in fig. 2.

In fig. 1, the resistance between C and E,

$\small\text{ \longrightarrow\sf{R_{CE}=6\,\Omega+1.5\,\Omega} }$

$\small\text{ \longrightarrow\sf{R_{CE}=7.5\,\Omega\quad\quad\dots(i)} }$

But in fig. 2,

$\small\text{ \longrightarrow\sf{R_{CE}=R_1\,||\,(R_2+R_3)} }$

$\small\text{ \longrightarrow\sf{R_{CE}=\dfrac{R_1(R_2+R_3)}{R_1+(R_2+R_3)}} }$

$\small\text{ \longrightarrow\sf{R_{CE}=\dfrac{R_1R_2+R_3R_1}{R_1+R_2+R_3}\quad\quad\dots(ii)} }$

From (i) and (ii),

$\small\text{ \longrightarrow\sf{\dfrac{R_1R_2+R_3R_1}{R_1+R_2+R_3}=7.5\,\Omega\quad\quad\dots(1)} }$

Similarly, we get the following.

$\small\text{ \longrightarrow\sf{\dfrac{R_2R_3+R_1R_2}{R_1+R_2+R_3}=3.5\,\Omega\quad\quad\dots(2)} }$

$\small\text{ \longrightarrow\sf{\dfrac{R_3R_1+R_2R_3}{R_1+R_2+R_3}=8\,\Omega\quad\quad\dots(3)} }$

Adding (1), (2) and (3) we get,

$\small\text{ \longrightarrow\sf{\dfrac{R_1R_2+R_2R_3+R_3R_1}{R_1+R_2+R_3}=\dfrac{7.5\,\Omega+3.5\,\Omega+8\,\Omega}{2}=9.5\,\Omega\quad\quad\dots(4)} }$

Then (4) - (1) gives,

$\small\text{ \longrightarrow\sf{\dfrac{R_2R_3}{R_1+R_2+R_3}=2\,\Omega\quad\quad\dots(5)} }$

(4) - (2) gives,

$\small\text{ \longrightarrow\sf{\dfrac{R_3R_1}{R_1+R_2+R_3}=6\,\Omega\quad\quad\dots(6)} }$

(4) - (3) gives,

$\small\text{ \longrightarrow\sf{\dfrac{R_1R_2}{R_1+R_2+R_3}=1.5\,\Omega\quad\quad\dots(7)} }$

Taking ratio between (5), (6) and (7) we get,

$\small\longrightarrow\sf{R_2R_3:R_3R_1:R_1R_2=2:6:1.5}$

$\small\text{ \longrightarrow\sf{\dfrac{R_1R_2R_3}{R_1}:\dfrac{R_1R_2R_3}{R_2}:\dfrac{R_1R_2R_3}{R_3}=2:6:1.5} }$

$\small\text{ \longrightarrow\sf{\dfrac{1}{R_1}:\dfrac{1}{R_2}:\dfrac{1}{R_3}=2:6:1.5} }$

$\small\text{ \longrightarrow\sf{R_1:R_2:R_3=\dfrac{1}{2}:\dfrac{1}{6}:\dfrac{1}{1.5}} }$

We can see that the ratio of resistances of resistors in delta connection is equal to the ratio of reciprocals of resistance of the resistor crossly connected to the corresponding branch in star connection, or vice versa.

$\small\text{ \longrightarrow\sf{R_1:R_2:R_3=\dfrac{1}{2}:\dfrac{1}{6}:\dfrac{2}{3}=3:1:4} }$

Take,

• $\small\sf{R_1=3k\,\Omega}$

• $\small\sf{R_2=k\,\Omega}$

• $\small\sf{R_3=4k\,\Omega}$

Then (5) becomes,

$\small\text{ \longrightarrow\sf{\dfrac{4k^2\,\Omega^2}{8k\,\Omega}=\dfrac{k}{2}\,\Omega=2\,\Omega} }$

$\small\longrightarrow\sf{k=4}$

Then,

• $\small\sf{R_1=12\,\Omega}$

• $\small\sf{R_2=4\,\Omega}$

• $\small\sf{R_3=16\,\Omega}$

Resistors of $\small\sf{3\,\Omega}$ and $\small\sf{R_1=12\,\Omega}$ resistances are in parallel so their equivalent resistance is $\small\text{ \sf{\dfrac{3\,\Omega\cdot12\,\Omega}{3\,\Omega+12\,\Omega}=2.4\,\Omega.} }$

This is connected in series with $\small\sf{R_3=16\,\Omega}$ resistor so their equivalent resistance is $\small\sf{2.4\,\Omega+16\,\Omega=18.4\,\Omega.}$

This is connected in parallel with $\small\sf{R_2=4\,\Omega}$ resistor so their equivalent resistance is $\small\text{ \sf{\dfrac{18.4\,\Omega\cdot4\,\Omega}{18.4\,\Omega+4\,\Omega}=\dfrac{23}{7}\,\Omega.} }$

This is the equivalent resistance of the whole circuit.

Hence current in the circuit is,

$\small\text{ \longrightarrow\sf{I=\dfrac{9\,V}{\left(\dfrac{23}{7}\,\Omega\right)}} }$

$\small\text{ \longrightarrow\underline{\underline{\sf{I=\dfrac{63}{23}\,A=2.74\,A}}} }$