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A circuit has a 2Ω resistor in series with a 4Ω resistor. What is the total resistance of the circuit?

Respuesta :

olatunbosunadeoke

Answer: 6Ω

Explanation:

Since the resistors are connected in series, the total resistance (Rtotal) of the circuit is the sum of each resistance.

i.e Rtotal = R1 + R2

R1 = 2Ω

R2 = 4Ω

Rtotal = ?

Rtotal = 2Ω + 4Ω

Rtotal = 6Ω

Thus, the total resistance of the circuit is 6Ω

ItzStarzMe

Answer :

  • Total resistance of the circuit is 6Ω

Explaination :

The circuit is of series which has a resistor of 2Ω and 4Ω respectively.

As we know that, if n resistances joined in series are [tex] \sf{R_1 \: + \:R_2 \: + \: R_3 \: ... \: R_n} [\tex] , the equivalent resistance is given as :

[tex]\boxed{ \red{\bf{R_s \: = \: R_1 \: + \:R_2 \: + \: R_3 \: ... \: R_n}}}[/tex]

Now,

  • [tex] \sf{R_1 \: = 2Ω} [/tex]
  • [tex] \sf{R_2 \: = 4Ω} [/tex]

Putting the values in it,

[tex] \implies \: \sf{R_s \: = \: 2Ω \: + \: 4Ω}[/tex]

[tex]\implies \: \bf{R_s \: = \: 6Ω}[/tex]

[tex] \underline{\bf{ \therefore Total \: resistance \: of \: the \: circuit \: is \: 6Ω }} [/tex]

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