A2 Physics 物理 出国英语2.4.3 Internal Resistance MS.pdfVIP

Internalresistancepracticequestionanswers

1.AsIrisesterminalvoltageofAfalls(1)

duetointernalresistanceofA/“lostvolts”(1)2

(i)‘Lostvolts’V=Ir=10.0×0.40=4.0V(1)

V=28.0–V=28.0–(20.0+4.0)=12.0(1)2

BA

(ii)P=VI=28.0×10.0=280W

Powersupplied=280W(1)1

(iii)P=I2r=10.02×0.40=40W

Powerwasted=40W(1)1

280

(iv)Eff==0.88=88%(1)2

320

Advantage=Renewablesupply/lastlonger(1)

Disadvantage=Dependsonillumination/highinternalresistance/large

arrayneededforpowerrequired(1)2

[10]

2.Explanation:.

I=E/r+R(1)1

AppropriateformulaforcellE9:

C9*D9ORRIOR1Ω×4A(1)1

AppropriateformulaforcellF11

D11*E11ORVIOR3A×6VORC11*D11*D11

ORRI2OR2Ω×(3A)2(1)1

Shortcircuitcurrent:

6A(1)1

Explanation:

randRinseries→potentialdivision(1)1

asR↑,rconstant