Answer:
Total resistance is 19.6 Ω for the circuit .
Explanation:
Given:
Six resistors of a circuit namely [tex]R_1[/tex] to [tex]R_6[/tex] .
Where
[tex]R_1[/tex] and [tex]R_2[/tex] are in series.
[tex]R_3[/tex] and [tex]R_4[/tex] are in parallel.
[tex]R_5[/tex] and [tex]R_6[/tex] are in parallel.
We know that continuous resistor which are in series are added up to find the equivalent resistance.
Similarly resistors which are arranged in parallel their equivalent resistance is [tex]\frac{R_3R_5}{R_3+R_5}[/tex] for the case of [tex]R_3[/tex] and [tex]R_5[/tex].
According to the question:
Total resistance = R(equivalent) :
⇒ [tex]R_(eq_) = R_1+R_2+(\frac{R_3R_4}{R_3+R_4} )+(\frac{R_5R_6}{R_5+R_6} )[/tex]
⇒ [tex]R_(eq_)=7+8+(\frac{4\times 6}{4+6} )+(\frac{3\times 8}{3+8} )[/tex]
⇒ [tex]R_(eq_)=7+8+(\frac{24}{10} )+(\frac{24}{11} )[/tex]
⇒ [tex]R_(eq_)=15+(2.4 )+(2.18)[/tex]
⇒ [tex]R_(eq_)=19.58[/tex] Ω
So the total resistance of the circuit depicted is 19.58 Ω approximated to nearest tenth that is 19.6 Ω
