Question 407: Current Electricity

Question

In the circuit shown, the current in the $$1\Omega $$ resistor is :

Accepted Answer

From $$KVL$$
$$ - 6 + 3{{m I}_1} + {m 1}\left( {{{m I}_1} - {{m I}_2}} ight) = 0$$
$$6 = 3{{m I}_1} + {{m I}_1} - {{m I}_2}$$
$$4{{m I}_1} - {{m I}_2} = 6\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 ight)$$
$$ - 9 + 2{{m I}_2} - \left( {{{m I}_1} - {{m I}_2}} ight) + 3{{m I}_2} = 0$$
$$ - {{m I}_1} + 6{{m I}_2} = 9\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 ight)$$
On solving $$\left( 1 ight)$$ and $$\left( 2 ight)$$
$${{m I}_1} = 0.13A$$
Direction $$Q$$ to $$P,$$ since $${{m I}_1} > {{m I}_2}.$$
Alternatively
$$Eq = {{{{{E_1}} \over {{r_1}}} + {{{E_2}} \over {{r_2}}}} \over {{1 \over {{r_1}}} + {1 \over {{r_2}}}}}$$
$$ = {{{6 \over 3} - {9 \over 5}} \over {{1 \over 3} + {1 \over 5}}} = {3 \over {8V}}$$
$$	herefore$$ $${m I} = {{{3 \over 8}} \over {{{15} \over 8} + 1}} = {3 \over {23}} = 0.13A$$
Considering potential at $$P$$ as $$0V$$ and at $$Q$$ as $$x$$ volt, then
$${{x - 6} \over 3} + {{x - 0} \over 1} + {{x + 9} \over 5} = 0$$
$$	herefore$$ $$x = {2 \over {23}}$$
$$	herefore$$ $$i = {{x - 0} \over 1} = {2 \over {23}} = 0.13A$$
From $$Q$$ to $$P$$

Answer

$$0.13$$ $$A,$$ from $$Q$$ to $$P$$

Answer

$$0.13$$ $$A$$, from $$P$$ to $$Q$$

Answer

$$1.3A$$ from $$P$$ to $$Q$$

Answer

$$0A$$

Current Electricity

In the circuit shown, the current in the $1\Omega $ resistor is :

Choose the correct answer: