Question

The diagram shows the arrangement of five different resistances connected to a battery of e.m.f. 1.8V. Calculate:

1. the total resistance of the circuit, and
2. the reading of ammeter A. 

Answer 1

(a) In circuit R₁ = 10 Ω

R₂ = 40 Ω are in parallel

`1/("R"_("p"_1)) = 1/10 + 1/40`

= `(4 + 1)/40`

= `5/40`

`"R"_("p"_2)` = 8 Ω

In circuit (ii) R₃ = 30 Ω, R₄ = 20 Ω, R₅ = 60 Ω

= `1/"R"_("p"_2) = 1/"R"_3 + 1/"R"_4 + 1/"R"_5` because they are in parallel

= `1/30 + 1/20 + 1/60`

= `(2 + 3 + 1)/60`

= `6/60`

`"R"_("p"_2) = 10  Ω`

Total resistance = `"R"_("p"_1) + "R"_("p"_2)`

= 8 Ω + 10 Ω

= 18 Ω

(b) R = `"V"/"I"`

∴ I = `"V"/"R"`

= `1.8/18`

= 0.1