Answer:
The final resistance is 5[tex]\Omega[/tex]
Explanation:
Refer to the following explanation I hope my answer is helpful. Thank you.Formulas Used:[tex]\rightarrow R_{eq} = \frac{R_1R_2}{R_1+R_2}[/tex] (In parallel)[tex]\rightarrow R_{eq} = R_1 + R_2[/tex] (In series)
The correct option is C
7.4
Ω
In the given circuit, resistances
6
and
4
are connected in parallel combination.
∴
their equivalent resistance
R
p
is given by:
1
=
+
⇒
×
10
24
2.4
Now, this equivalant resistance is connected in series combination with
5
resistance.
the effective resistance of the circuit is:
e
f
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Verified answer
Answer:
The final resistance is 5[tex]\Omega[/tex]
Explanation:
Refer to the following explanation I hope my answer is helpful. Thank you.
Formulas Used:
[tex]\rightarrow R_{eq} = \frac{R_1R_2}{R_1+R_2}[/tex] (In parallel)
[tex]\rightarrow R_{eq} = R_1 + R_2[/tex] (In series)
The correct option is C
7.4
Ω
In the given circuit, resistances
6
Ω
and
4
Ω
are connected in parallel combination.
∴
their equivalent resistance
R
p
is given by:
1
R
p
=
1
6
Ω
+
1
4
Ω
⇒
1
R
p
=
6
Ω
+
4
Ω
6
Ω
×
4
Ω
=
10
24
Ω
⇒
R
p
=
2.4
Ω
Now, this equivalant resistance is connected in series combination with
5
Ω
resistance.
∴
the effective resistance of the circuit is:
R
e
f
f
=
5
Ω
+
2.4
Ω
=
7.4
Ω
please mark me as brainliest..