#### To determine

**To find:** The rate of change of the total resistance R of the circuit.

#### Answer

The rate of change of the total resistance R of the circuit is dRdt=107810≈0.132 Ω/s.

#### Explanation

**Given:**

The rate of change of the resistance R1 is 0.3Ω/s .

The rate of change of the resistance R2 is 0.2Ω/s .

**Formula used:**

(1) Chain rule: dydx=dydu⋅dudx.

(2) ddx[fg]=gdfdx−fdgdx[g]2.

**Calculation:**

Since the resistance R1 and R2 are connected in the parallel combination and R is the total equivalent resistance of the given circuit.

Therefore,

1R=1R1+1R2 .

Now first find R.

1R=180+11001R=5+44001R=9400R=4009 Ω

Consider the resistance R1 and R2 changes with time t, therefore the total resistance R is also changes with the time t.

Since dR1dt=0.3 Ω/s and dR2dt=0.2 Ω/s .

Find dRdt when R1=80 Ω and R2=100 Ω

Differentiate 1R=1R1+1R2 with respect to the time t.

ddt[1R]=ddt[1R1+1R2]ddR[1R]dRdt=ddR1[1R1]dR1dt+ddR2[1R2]dR2dt [Qdydx=dydu⋅dudx]−1R2dRdt=−1R12dR1dt+(−1R22)dR2dt [Qddx[fg]=gdfdx−fdgdx[g]2]1R2dRdt=1R12dR1dt+(1R22)dR2dt

On further simplification the rate of change of the total resistance R as follows.

dRdt=R2[1R12dR1dt+(1R22)dR2dt]

Substitutes R1=80 ,R2=100 ,R=4009 and dR1dt=0.3 and dR2dt=0.2 in dRdt

dRdt=400922[1802(0.3)+11002(0.2)]=400292[0.36400+0.210000]=1600009[0.36400+0.210000]=16009[0.364+0.2100] Ω/s

On further simplification the rate of change of the total resistance R as follows.

dRdt=160081[0.00468+0.002] =160081[0.00668]=10.66881=0.132 Ω/s

Therefore, the rate of change of the total resistance R of the circuit is dRdt=107810≈0.132 Ω/s.