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In the circuit shown below, the Norton equivalent current in amperes with respect to the terminals $\mathrm{P}$ and $\mathrm{Q}$ is

  1. $6.4-\mathrm{j} 4.8$
  2. $6.56-\mathrm{j} 7.87$
  3. $10+\mathrm{j} 0$
  4. $16+\mathrm{j} 0$
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