Question

Draw a schematic diagram of a circuit consisting of a battery of five 2 V cells, a 5 ohm resistor, a 10 ohm resistor, a 15 ohm resistor, an ammeter and a plug key, all connected in series. Calculate the electric current passing through the above circuit when the key is closed..
Please help ASAP..​

Answer 1

Answer:

Resistance = 5 + 10 + 15 = 30

Explanation:

Current= 0.33 Amperes

Answer 2

Answer:

• Current = 0.33 A

Explanation:

• For diagram refer the attachment.

It is given that five cells of 2V are connected in series, so total voltage of the battery:

$\dashrightarrow \: \: \sf V = 2 \times 5 = 10 V$

Three resistor of 5$\Omega$, 10$\Omega$, 15$\Omega$ are connected in Series, so the net resistance:

$\dashrightarrow \: \: \sf R_{n} = R_{1} + R_{2} + R_{3}$

$\dashrightarrow \: \: \sf R = 5 + 10 + 15$

${ \pink{\dashrightarrow \sf \: \: { \underbrace{R = 30 \: \Omega}}}}$

According to ohm's law:

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

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

On substituting resultant voltage (V) as 10 V and resultant resistant, as 30 ${\pmb{\sf{\Omega}}}$ we get:

$\dashrightarrow \sf \: \: I = \dfrac{10V}{30\Omega}$

${ \pink{\dashrightarrow \sf \: \: { \underbrace{I = 0.33 A}}}}$

$\therefore$The electric current passing through the above circuit when the key is closed will be 0.33 A