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Use the circuit in figure 7 below to answer the questions that follow (i) Calculate resistance in the circuit. (ii) Determine the current flowing through the 5 Ω resistor.

Physics

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Huu hapa ni ufumbuzi wa maswali yaliyoonekana kwenye picha.

Swali (b):
Taja sababu mbili zinazoathiri upinzani wa kondakta ya metali. (alama 1)

  • Urefu wa kondakta.
  • Eneo la msalaba (cross-sectional area) la kondakta.
  • Aina ya nyenzo (resistivity) ya kondakta.
  • Joto la kondakta.

Swali (c):
Tumia mzunguko katika kielelezo cha 7 hapa chini kujibu maswali yanayofuata.

Swali (c)(i):
Kokotoa jumla ya upinzani katika mzunguko.

  1. Kokotoa upinzani wa kwanza wa sambamba (Rp1R_{p1}).

    Kuna vipingamizi vitatu vilivyounganishwa sambamba: 10 Ω10 \, \Omega, 20 Ω20 \, \Omega, na 20 Ω20 \, \Omega.

    1Rp1=110 Ω+120 Ω+120 Ω\frac{1}{R_{p1}} = \frac{1}{10 \, \Omega} + \frac{1}{20 \, \Omega} + \frac{1}{20 \, \Omega} 1Rp1=220 Ω+120 Ω+120 Ω=420 Ω=15 Ω\frac{1}{R_{p1}} = \frac{2}{20 \, \Omega} + \frac{1}{20 \, \Omega} + \frac{1}{20 \, \Omega} = \frac{4}{20 \, \Omega} = \frac{1}{5 \, \Omega} Rp1=5 ΩR_{p1} = 5 \, \Omega
  2. Kokotoa upinzani wa pili wa sambamba (Rp2R_{p2}).

    Kuna vipingamizi viwili vilivyounganishwa sambamba: 10 Ω10 \, \Omega na 5 Ω5 \, \Omega.

    1Rp2=110 Ω+15 Ω\frac{1}{R_{p2}} = \frac{1}{10 \, \Omega} + \frac{1}{5 \, \Omega} 1Rp2=110 Ω+210 Ω=310 Ω\frac{1}{R_{p2}} = \frac{1}{10 \, \Omega} + \frac{2}{10 \, \Omega} = \frac{3}{10 \, \Omega} Rp2=103 ΩR_{p2} = \frac{10}{3} \, \Omega
  3. Kokotoa jumla ya upinzani (RtotalR_{total}).

    Vipingamizi Rp1R_{p1}, 3 Ω3 \, \Omega, na Rp2R_{p2} vimeunganishwa kwa mfululizo.

    Rtotal=Rp1+3 Ω+Rp2R_{total} = R_{p1} + 3 \, \Omega + R_{p2} Rtotal=5 Ω+3 Ω+103 ΩR_{total} = 5 \, \Omega + 3 \, \Omega + \frac{10}{3} \, \Omega Rtotal=8 Ω+103 ΩR_{total} = 8 \, \Omega + \frac{10}{3} \, \Omega

    Ili kuongeza, tunahitaji denominator ya kawaida:

    Rtotal=243 Ω+103 ΩR_{total} = \frac{24}{3} \, \Omega + \frac{10}{3} \, \Omega Rtotal=343 ΩR_{total} = \frac{34}{3} \, \Omega Rtotal≈11.33 ΩR_{total} \approx 11.33 \, \Omega

    Jumla ya upinzani katika mzunguko ni 11.33 \Omega\boxed{\text{11.33 \Omega}}.

    Swali (c)(ii):
    Tambua mkondo wa umeme unaopita kwenye kipingamizi cha 5 Ω5 \, \Omega.

  4. Kokotoa jumla ya mkondo wa umeme (ItotalI_{total}) kutoka chanzo.

    Tumia Sheria ya Ohm, I=VRI = \frac{V}{R}. Voltage ya chanzo ni 20 V20 \, V.

    Itotal=20 V343 ΩI_{total} = \frac{20 \, V}{\frac{34}{3} \, \Omega} Itotal=20×334 A=6034 A=3017 AI_{total} = \frac{20 \times 3}{34} \, A = \frac{60}{34} \, A = \frac{30}{17} \, A
  5. Kokotoa voltage kwenye kipingamizi cha sambamba cha Rp2R_{p2}.

    Mkondo wa jumla unapita kwenye Rp2R_{p2}.

    Vp2=Itotal×Rp2V_{p2} = I_{total} \times R_{p2} Vp2=3017 A×103 ΩV_{p2} = \frac{30}{17} \, A \times \frac{10}{3} \, \Omega Vp2=30051 V=10017 VV_{p2} = \frac{300}{51} \, V = \frac{100}{17} \, V
  6. Kokotoa mkondo wa umeme unaopita kwenye kipingamizi cha 5 Ω5 \, \Omega.

    Kwa kuwa vipingamizi vya 10 Ω10 \, \Omega na 5 Ω5 \, \Omega vimeunganishwa sambamba, voltage kwenye kila kipingamizi ni Vp2V_{p2}.

    I5Ω=Vp25 ΩI_{5\Omega} = \frac{V_{p2}}{5 \, \Omega} I5Ω=10017 V5 ΩI_{5\Omega} = \frac{\frac{100}{17} \, V}{5 \, \Omega} I5Ω=10017×5 A=2017 AI_{5\Omega} = \frac{100}{17 \times 5} \, A = \frac{20}{17} \, A I5Ω≈1.18 AI_{5\Omega} \approx 1.18 \, A

    Mkondo wa umeme unaopita kwenye kipingamizi cha 5 Ω5 \, \Omega ni 1.18 A\boxed{\text{1.18 A}}.

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