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Tnoda · Answer 1 · 2021-08-03T21:28:55+0000

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
$\sf{R_1 \: + \:R_2 \: + \: R_3 \: ... \: R_n} [\tex] , the equivalent resistance is given as : </p><p>[tex]\boxed{ \red{\bf{R_s \: = \: R_1 \: + \:R_2 \: + \: R_3 \: ... \: R_n}}}$

Now,

$\sf{R_1 \: = 2Ω}$
$\sf{R_2 \: = 4Ω}$

Putting the values in it,

$\implies \: \sf{R_s \: = \: 2Ω \: + \: 4Ω}$

$\implies \: \bf{R_s \: = \: 6Ω}$

$\underline{\bf{ \therefore Total \: resistance \: of \: the \: circuit \: is \: 6Ω }}$

Neel Bhanushali · Answer 2 · 2021-08-07T12:41:12+0000

Answer: 6Ω

Step-by-step 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

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