Tiga buah lampu listrik dengan nilai hambatan R¹=7𝞨 R²=3𝞨 R³=21𝞨 lampu 1 dan lampu 2 dirangkai paralel selanjutnua diserikan dengan R³.Rangkaian itu dialiri listrik sebesar 40v,maka hitunglah: a.Rpengganti b.Rtotal c.Itotal d.VR³ e.VRP f.IR²

<p>R₁ = 7 𝞨</p><p>R₂ = 3 𝞨</p><p>R₃ = 21 𝞨</p><p>V = 40 volt</p><p>&nbsp;</p><p>a. Mencari R Pengganti</p><p>&nbsp; &nbsp; &nbsp;R₁ dan R₂ dirangkai Paralel</p><p>&nbsp; &nbsp; &nbsp;1/R_p = 1/R₁ + 1/R₂</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= 1/7 + 1/3</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= 3/21 + 7/21</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= 10/21</p><p>&nbsp; &nbsp; &nbsp; R_p = 21/10 = 2,1 𝞨</p><p>&nbsp; &nbsp; &nbsp;Besar hambatan pengganti adalah 2,1 𝞨</p><p>&nbsp;</p><p>b. Mencari R Total</p><p>&nbsp; &nbsp; &nbsp;R_total = R_p + R₃</p><p>_{&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;} &nbsp; &nbsp;= 2,1 + 21</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= 23,1 𝞨</p><p>&nbsp;</p><p>c. Mencari I_total</p><p>&nbsp; &nbsp; &nbsp;I_total&nbsp; = V/R_total</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= 40/23,1</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= 1,73 Ampere</p><p>&nbsp;</p><p>d. Tegangan pada R₃</p><p>&nbsp; &nbsp; &nbsp; V = I_total x R₃</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = 1,73 x 21</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = 36,33 volt</p><p>&nbsp;</p><p>e. Tegangan pada R_p</p><p>&nbsp; &nbsp; &nbsp; V_p = I_total&nbsp; x R_p</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= 1,73 x 2,1</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= 3,633 volt</p><p>&nbsp;</p><p>f. Kuat arus pada R₃</p><p>&nbsp; &nbsp; &nbsp; &nbsp;I₃ = V_p/R₃</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= 3,633 / 21</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= 0,173 Ampere</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;</p>