Question

The resistance is 6 Ohm is connected in series with another resistance 4 Ohm. a potential difference of 20v is applied across the combination. Calculate the current through circuit & potential difference across its circuit?​

Answer 1

${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}}$

$\:\:\:\:\bullet\:\:\:\sf{Two \:resistors \:each \:of \:6 \Omega\: and \:4 \Omega \:are\: connected\: in \:series}$

$\:\:\:\:\bullet\:\:\:\sf{Potential \:difference \:(V) = 20\: volts}$

${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}}$

$\:\:\:\:\bullet\:\:\:\sf{Potential\: difference}$

$\:\:\:\:\bullet\:\:\:\sf{Current\: through\: the \:circuit}$

${\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}}$

☯ Firstly, let's find net resistance

$\dashrightarrow\:\: \sf{ R_{net} = R_1 + R_2............R_n}$

$\dashrightarrow\:\: \sf{R_{net} = 6 \Omega + 4 \Omega }$

$\dashrightarrow\:\: {\boxed{\sf{R_{net} = 10 \Omega }}}$

☯ Now, find the current in the circuit

$\dashrightarrow\:\: \sf{ V=IR }$

$\dashrightarrow\:\: \sf{I = \dfrac{V}{R} }$

$\dashrightarrow\:\: \sf{I = \dfrac{\cancel{20}}{\cancel{10}} }$

$\dashrightarrow\:\: {\boxed{\sf{ I = 2 \: A}}}$

• Electric current through circuit is $\bf{2A}$ & potential difference be same i.e., $\bf{20V}$

Answer 2

$\huge\mathcal{\underline{\color{purple} Answer}}$

Using Ohm’s law, we find the current I = V/Req

So we can start by finding Req, in series connected resistances, Req = R1 + R2 + … + Rn; therefore, Req = 6 + 4 = 10 Ohms.

We find the current I = V/Req = 20 / 10 = 2 Amperes.

The potential difference on each element is found by using the same Ohm’s, V = I * R; using “I” as found previously since in a series connected circuit, the current is the same in all the elements,

V = I * R

= 2 * 6

= 12 Volts.

Hope it helps uh !!