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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..​

Sagot :

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:

[tex]\dashrightarrow \: \: \sf V = 2 \times 5 = 10 V [/tex]

Three resistor of 5[tex]\Omega[/tex], 10[tex]\Omega[/tex], 15[tex]\Omega[/tex] are connected in Series, so the net resistance:

[tex]\dashrightarrow \: \: \sf R_{n} = R_{1} + R_{2} + R_{3}[/tex]

[tex]\dashrightarrow \: \: \sf R = 5 + 10 + 15 [/tex]

[tex]{ \pink{\dashrightarrow \sf \: \: { \underbrace{R = 30 \: \Omega}}}}[/tex]

According to ohm's law:

[tex]\dashrightarrow \sf\: \: V = IR [/tex]

[tex]\dashrightarrow \sf \: \: I = \dfrac{V}{R}[/tex]

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

[tex]\dashrightarrow \sf \: \: I = \dfrac{10V}{30\Omega}[/tex]

[tex]{ \pink{\dashrightarrow \sf \: \: { \underbrace{I = 0.33 A}}}}[/tex]

[tex]\therefore[/tex]The electric current passing through the above circuit when the key is closed will be 0.33 A

View image GlitterSparks

Answer:

Resistance = 5 + 10 + 15 = 30

Explanation:

Current= 0.33 Amperes

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