HomeLearn Fast select Course
Show Unit/ChapterChange UnitChange SubjectLast Menu
NEET-XII-Physics

32: Electric Current in Conductors

with Solutions - page 5
Qstn# iv-20 Prvs-QstnNext-Qstn
  • #20
    Find the value of i1/i2 in the figure (32-E3) if (a) R = 0.1 Ω (b) R = 1 Ω and (c) R = 10 Ω. Note from your answers that in order to get more current from a combination of two batteries, they should be joined in parallel if the external resistance is small and in series if the external resistance is large, compared to the internal resistance.
    Figure 32-E3 (a) R = 0.1 Ω (b) R = 1 Ω and (c) R = 10 Ω. Note from your answers that in order to get more current from a combination of two batteries, they should be joined in parallel if the external resistance is small and in series if the external resistance is large, compared to the internal resistance.
    Figure 32-E3
    Ans : (a) For R = 0.1 Ω:
    Applying KVL in the given circuit, we get:

    `` 0.1{i}_{1}+1{i}_{1}-6+1{i}_{1}-6=0``
    `` \Rightarrow 0.1{i}_{1}+1{i}_{1}+1{i}_{1}=12``
    `` \Rightarrow {i}_{1}=\frac{12}{\left(2.1\right)}=5.71\,\mathrm{\,A\,}``
    `` ``
    Now, consider the given circuit.

    Applying KVL in the loop ABCDA, we get:
    `` 0.1{i}_{2}+1i-6=0``
    `` \Rightarrow 0.1{i}_{2}+i=6``
    `` \Rightarrow i=6-0.1{i}_{2}``
    Applying KVL in ADEFA, we get:
    `` i-6+6-\left({i}_{2}-i\right)1=0``
    `` \Rightarrow i-{i}_{2}+i=0``
    `` \Rightarrow 2\,\mathrm{\,i\,}-{i}_{2}=0``
    `` \Rightarrow 2\left[6-0.1{i}_{2}\right]-{i}_{2}=0``
    `` \Rightarrow {i}_{2}=10\,\mathrm{\,A\,}``
    `` \therefore \frac{{i}_{1}}{{i}_{2}}=0.571`` (b) For R = 1 Ω:
    Applying KVL in the circuit given in figure 1, we get:
    `` 1{i}_{1}+1.{i}_{1}-6+{i}_{1}-6=0``
    `` \Rightarrow 3{i}_{1}=12``
    `` \Rightarrow {i}_{1}=4``
    `` ``
    Now, for figure 2:
    Applying KVL in ABCDA, we get:
    `` {i}_{2}+i-6=0``
    `` \Rightarrow {i}_{2}+i=6``
    Applying KVL in ADEFA, we get:
    `` i-6+6-\left({i}_{2}-i\right)1=0``
    `` \Rightarrow i-{i}_{2}+i=0``
    `` \Rightarrow 2i-{i}_{2}=0``
    `` \Rightarrow 2\left[6-{i}_{2}\right]-{i}_{2}=0``
    `` \Rightarrow 12-3{i}_{2}=0``
    `` \Rightarrow {i}_{2}=4\,\mathrm{\,A\,}``
    `` \therefore \frac{{i}_{1}}{{i}_{2}}=1`` (c) For R = 10 Ω:
    Applying KVL in the circuit given in figure 1, we get:
    `` 10{i}_{1}+1{i}_{1}-6+1{i}_{1}-6=0``
    `` \Rightarrow 12{i}_{1}=12``
    `` \Rightarrow {i}_{1}=1``
    `` ``
    Now, for figure 2:
    Applying KVL in ABCDA, we get:
    `` 10{i}_{2}+i-6=0``
    `` \Rightarrow i=6-10{i}_{2}``
    Applying KVL in ADEFA, we get:
    `` i-6+6-\left({i}_{2}-i\right)1=0``
    `` \Rightarrow i-{i}_{2}+i=0``
    `` \Rightarrow 2i-{i}_{2}=0``
    `` \Rightarrow 2\left[6-10{i}_{2}\right]-{i}_{2}=0``
    `` \Rightarrow 12-21{i}_{2}=0``
    `` \Rightarrow {i}_{2}=0.57\,\mathrm{\,A\,}``
    `` \therefore \frac{{i}_{1}}{{i}_{2}}=1.75``
    Page No 199: (a) For R = 0.1 Ω:
    Applying KVL in the given circuit, we get:

    `` 0.1{i}_{1}+1{i}_{1}-6+1{i}_{1}-6=0``
    `` \Rightarrow 0.1{i}_{1}+1{i}_{1}+1{i}_{1}=12``
    `` \Rightarrow {i}_{1}=\frac{12}{\left(2.1\right)}=5.71\,\mathrm{\,A\,}``
    `` ``
    Now, consider the given circuit.

    Applying KVL in the loop ABCDA, we get:
    `` 0.1{i}_{2}+1i-6=0``
    `` \Rightarrow 0.1{i}_{2}+i=6``
    `` \Rightarrow i=6-0.1{i}_{2}``
    Applying KVL in ADEFA, we get:
    `` i-6+6-\left({i}_{2}-i\right)1=0``
    `` \Rightarrow i-{i}_{2}+i=0``
    `` \Rightarrow 2\,\mathrm{\,i\,}-{i}_{2}=0``
    `` \Rightarrow 2\left[6-0.1{i}_{2}\right]-{i}_{2}=0``
    `` \Rightarrow {i}_{2}=10\,\mathrm{\,A\,}``
    `` \therefore \frac{{i}_{1}}{{i}_{2}}=0.571`` (b) For R = 1 Ω:
    Applying KVL in the circuit given in figure 1, we get:
    `` 1{i}_{1}+1.{i}_{1}-6+{i}_{1}-6=0``
    `` \Rightarrow 3{i}_{1}=12``
    `` \Rightarrow {i}_{1}=4``
    `` ``
    Now, for figure 2:
    Applying KVL in ABCDA, we get:
    `` {i}_{2}+i-6=0``
    `` \Rightarrow {i}_{2}+i=6``
    Applying KVL in ADEFA, we get:
    `` i-6+6-\left({i}_{2}-i\right)1=0``
    `` \Rightarrow i-{i}_{2}+i=0``
    `` \Rightarrow 2i-{i}_{2}=0``
    `` \Rightarrow 2\left[6-{i}_{2}\right]-{i}_{2}=0``
    `` \Rightarrow 12-3{i}_{2}=0``
    `` \Rightarrow {i}_{2}=4\,\mathrm{\,A\,}``
    `` \therefore \frac{{i}_{1}}{{i}_{2}}=1`` (c) For R = 10 Ω:
    Applying KVL in the circuit given in figure 1, we get:
    `` 10{i}_{1}+1{i}_{1}-6+1{i}_{1}-6=0``
    `` \Rightarrow 12{i}_{1}=12``
    `` \Rightarrow {i}_{1}=1``
    `` ``
    Now, for figure 2:
    Applying KVL in ABCDA, we get:
    `` 10{i}_{2}+i-6=0``
    `` \Rightarrow i=6-10{i}_{2}``
    Applying KVL in ADEFA, we get:
    `` i-6+6-\left({i}_{2}-i\right)1=0``
    `` \Rightarrow i-{i}_{2}+i=0``
    `` \Rightarrow 2i-{i}_{2}=0``
    `` \Rightarrow 2\left[6-10{i}_{2}\right]-{i}_{2}=0``
    `` \Rightarrow 12-21{i}_{2}=0``
    `` \Rightarrow {i}_{2}=0.57\,\mathrm{\,A\,}``
    `` \therefore \frac{{i}_{1}}{{i}_{2}}=1.75``
    Page No 199:
  • #20-a
    R = 0.1 Ω
    Ans : For R = 0.1 Ω:
    Applying KVL in the given circuit, we get:

    `` 0.1{i}_{1}+1{i}_{1}-6+1{i}_{1}-6=0``
    `` \Rightarrow 0.1{i}_{1}+1{i}_{1}+1{i}_{1}=12``
    `` \Rightarrow {i}_{1}=\frac{12}{\left(2.1\right)}=5.71\,\mathrm{\,A\,}``
    `` ``
    Now, consider the given circuit.

    Applying KVL in the loop ABCDA, we get:
    `` 0.1{i}_{2}+1i-6=0``
    `` \Rightarrow 0.1{i}_{2}+i=6``
    `` \Rightarrow i=6-0.1{i}_{2}``
    Applying KVL in ADEFA, we get:
    `` i-6+6-\left({i}_{2}-i\right)1=0``
    `` \Rightarrow i-{i}_{2}+i=0``
    `` \Rightarrow 2\,\mathrm{\,i\,}-{i}_{2}=0``
    `` \Rightarrow 2\left[6-0.1{i}_{2}\right]-{i}_{2}=0``
    `` \Rightarrow {i}_{2}=10\,\mathrm{\,A\,}``
    `` \therefore \frac{{i}_{1}}{{i}_{2}}=0.571``
  • #20-b
    R = 1 Ω and
    Ans : For R = 1 Ω:
    Applying KVL in the circuit given in figure 1, we get:
    `` 1{i}_{1}+1.{i}_{1}-6+{i}_{1}-6=0``
    `` \Rightarrow 3{i}_{1}=12``
    `` \Rightarrow {i}_{1}=4``
    `` ``
    Now, for figure 2:
    Applying KVL in ABCDA, we get:
    `` {i}_{2}+i-6=0``
    `` \Rightarrow {i}_{2}+i=6``
    Applying KVL in ADEFA, we get:
    `` i-6+6-\left({i}_{2}-i\right)1=0``
    `` \Rightarrow i-{i}_{2}+i=0``
    `` \Rightarrow 2i-{i}_{2}=0``
    `` \Rightarrow 2\left[6-{i}_{2}\right]-{i}_{2}=0``
    `` \Rightarrow 12-3{i}_{2}=0``
    `` \Rightarrow {i}_{2}=4\,\mathrm{\,A\,}``
    `` \therefore \frac{{i}_{1}}{{i}_{2}}=1``
  • #20-c
    R = 10 Ω. Note from your answers that in order to get more current from a combination of two batteries, they should be joined in parallel if the external resistance is small and in series if the external resistance is large, compared to the internal resistance.
    Figure 32-E3
    Ans : For R = 10 Ω:
    Applying KVL in the circuit given in figure 1, we get:
    `` 10{i}_{1}+1{i}_{1}-6+1{i}_{1}-6=0``
    `` \Rightarrow 12{i}_{1}=12``
    `` \Rightarrow {i}_{1}=1``
    `` ``
    Now, for figure 2:
    Applying KVL in ABCDA, we get:
    `` 10{i}_{2}+i-6=0``
    `` \Rightarrow i=6-10{i}_{2}``
    Applying KVL in ADEFA, we get:
    `` i-6+6-\left({i}_{2}-i\right)1=0``
    `` \Rightarrow i-{i}_{2}+i=0``
    `` \Rightarrow 2i-{i}_{2}=0``
    `` \Rightarrow 2\left[6-10{i}_{2}\right]-{i}_{2}=0``
    `` \Rightarrow 12-21{i}_{2}=0``
    `` \Rightarrow {i}_{2}=0.57\,\mathrm{\,A\,}``
    `` \therefore \frac{{i}_{1}}{{i}_{2}}=1.75``
    Page No 199: