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

Resistance = 5 + 10 + 15 = 30

Step-by-step explanation:

Current= 0.33 Amperes

Answer 2

• Current = 0.33 A

Step-by-step 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 * 5 = <strong>10 V</strong>`

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 = (V)/(R)`

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

`\dashrightarrow \sf \: \: I = (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