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Kama vipengee vilivyoonyeshwa kwenye michoro yote miwili vimeunganishwa kwa usahihi, tafadhali kokotoa kipingamizi kinachofaa kwa kila mchoro. Kwa kila mchoro, onyesha hatua zako zote za hesabu, pamoj

Physics

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Hapa kuna hesabu ya vipingamizi vinavyofaa kwa kila mchoro:

1. Mchoro wa 1:

  1. Tambua vipingamizi vilivyounganishwa kwa mfululizo katika tawi la chini la sambamba.

    Rchini=1 Ω+2 Ω+1 Ω=4 ΩR_{chini} = 1\,\Omega + 2\,\Omega + 1\,\Omega = 4\,\Omega
  2. Kokotoa kipingamizi kinachofaa cha vipingamizi vilivyounganishwa kwa sambamba (tawi la juu la 4 Ω4\,\Omega na tawi la chini la 4 Ω4\,\Omega).

    Rsambamba=4 Ω×4 Ω4 Ω+4 Ω=16 Ω28 Ω=2 ΩR_{sambamba} = \frac{4\,\Omega \times 4\,\Omega}{4\,\Omega + 4\,\Omega} = \frac{16\,\Omega^2}{8\,\Omega} = 2\,\Omega
  3. Kokotoa kipingamizi kinachofaa cha jumla kwa kuongeza kipingamizi cha mwanzo cha 2 Ω2\,\Omega na kipingamizi cha sambamba.

    Rjumla=2 Ω+2 Ω=4 ΩR_{jumla} = 2\,\Omega + 2\,\Omega = 4\,\Omega

    Kipingamizi kinachofaa kwa mchoro wa 1 ni: 4 Ω\boxed{\text{4}\,\Omega}

    2. Mchoro wa 2:

  4. Kokotoa kipingamizi cha jumla cha tawi la juu la sambamba.

    Rjuu=50 Ω+200 Ω=250 ΩR_{juu} = 50\,\Omega + 200\,\Omega = 250\,\Omega
  5. Kokotoa kipingamizi cha jumla cha tawi la chini la sambamba.

    Rchini=300 Ω+100 Ω=400 ΩR_{chini} = 300\,\Omega + 100\,\Omega = 400\,\Omega
  6. Kokotoa kipingamizi kinachofaa cha vipingamizi vilivyounganishwa kwa sambamba.

    Rsambamba=250 Ω×400 Ω250 Ω+400 Ω=100000 Ω2650 Ω=1000065 Ω=200013 ΩR_{sambamba} = \frac{250\,\Omega \times 400\,\Omega}{250\,\Omega + 400\,\Omega} = \frac{100000\,\Omega^2}{650\,\Omega} = \frac{10000}{65}\,\Omega = \frac{2000}{13}\,\Omega
  7. Kokotoa kipingamizi kinachofaa cha jumla kwa kuongeza kipingamizi cha sambamba na kipingamizi cha mwisho cha 250 Ω250\,\Omega.

    Rjumla=200013 Ω+250 Ω=2000+(250×13)13 Ω=2000+325013 Ω=525013 ΩR_{jumla} = \frac{2000}{13}\,\Omega + 250\,\Omega = \frac{2000 + (250 \times 13)}{13}\,\Omega = \frac{2000 + 3250}{13}\,\Omega = \frac{5250}{13}\,\Omega

    Kipingamizi kinachofaa kwa mchoro wa 2 ni: 525013 Ω≈403.85 Ω\boxed{\frac{5250}{13}\,\Omega \approx 403.85\,\Omega}

    3. Mchoro wa 3:

  8. Kokotoa kipingamizi cha jumla cha tawi la chini la sambamba.

    Rchini=3 Ω+4 Ω=7 ΩR_{chini} = 3\,\Omega + 4\,\Omega = 7\,\Omega
  9. Kokotoa kipingamizi kinachofaa cha vipingamizi vilivyounganishwa kwa sambamba (tawi la juu la 6 Ω6\,\Omega na tawi la chini la 7 Ω7\,\Omega).

    Rsambamba=6 Ω×7 Ω6 Ω+7 Ω=42 Ω213 Ω=4213 ΩR_{sambamba} = \frac{6\,\Omega \times 7\,\Omega}{6\,\Omega + 7\,\Omega} = \frac{42\,\Omega^2}{13\,\Omega} = \frac{42}{13}\,\Omega
  10. Kokotoa kipingamizi kinachofaa cha jumla kwa kuongeza vipingamizi vyote vilivyounganishwa kwa mfululizo.

    Rjumla=5 Ω+4213 Ω+2 Ω=7 Ω+4213 Ω=(7×13)+4213 Ω=91+4213 Ω=13313 ΩR_{jumla} = 5\,\Omega + \frac{42}{13}\,\Omega + 2\,\Omega = 7\,\Omega + \frac{42}{13}\,\Omega = \frac{(7 \times 13) + 42}{13}\,\Omega = \frac{91 + 42}{13}\,\Omega = \frac{133}{13}\,\Omega

    Kipingamizi kinachofaa kwa mchoro wa 3 ni: 13313 Ω≈10.23 Ω\boxed{\frac{133}{13}\,\Omega \approx 10.23\,\Omega}

    4. Mchoro wa 4:

  11. Kokotoa kipingamizi cha jumla cha tawi la chini la sambamba.

    Rchini=4 Ω+1 Ω=5 ΩR_{chini} = 4\,\Omega + 1\,\Omega = 5\,\Omega
  12. Kokotoa kipingamizi kinachofaa cha vipingamizi vilivyounganishwa kwa sambamba (tawi la juu la 1 Ω1\,\Omega na tawi la chini la 5 Ω5\,\Omega).

    Rsambamba=1 Ω×5 Ω1 Ω+5 Ω=5 Ω26 Ω=56 ΩR_{sambamba} = \frac{1\,\Omega \times 5\,\Omega}{1\,\Omega + 5\,\Omega} = \frac{5\,\Omega^2}{6\,\Omega} = \frac{5}{6}\,\Omega
  13. Kokotoa kipingamizi kinachofaa cha jumla kwa kuongeza vipingamizi vyote vilivyounganishwa kwa mfululizo.

    Rjumla=2 Ω+56 Ω+2 Ω=4 Ω+56 Ω=(4×6)+56 Ω=24+56 Ω=296 ΩR_{jumla} = 2\,\Omega + \frac{5}{6}\,\Omega + 2\,\Omega = 4\,\Omega + \frac{5}{6}\,\Omega = \frac{(4 \times 6) + 5}{6}\,\Omega = \frac{24 + 5}{6}\,\Omega = \frac{29}{6}\,\Omega

    Kipingamizi kinachofaa kwa mchoro wa 4 ni: 296 Ω≈4.83 Ω\boxed{\frac{29}{6}\,\Omega \approx 4.83\,\Omega}

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